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Provide a correct Python 3 solution for this coding contest problem. problem There are the following games. N characters are lined up in a vertical row. The color of these characters is red, blue, or yellow, and in the initial state, four or more characters of the same color are not lined up in a row. The player can select a character at a certain position and change it to another color. By this operation, if four or more characters of the same color are lined up in a row, those characters will disappear. When four or more characters of the same color are lined up in a row due to the disappearance of the characters, those characters also disappear, and this chain continues until there are no more places where four or more characters of the same color are lined up in a row. .. The purpose of this game is to reduce the number of characters remaining without disappearing. For example, if the color of the sixth character from the top is changed from yellow to blue in the state at the left end of the figure below, five blue characters will disappear in a row, and finally three characters will remain without disappearing. <image> Given the color sequence of N characters in the initial state, create a program that finds the minimum value M of the number of characters that remain without disappearing when the color of the character is changed in only one place. input The input consists of multiple datasets. Each dataset is given in the following format. The first line consists of only the number of characters N (1 ≀ N ≀ 10000). The following N lines contain one of 1, 2, and 3 integers, and the i + 1st line (1 ≀ i ≀ N) represents the color of the i-th character from the top in the initial state (1). Is red, 2 is blue, and 3 is yellow). When N is 0, it indicates the end of input. The number of datasets does not exceed 5. output For each dataset, output the minimum value M of the number of characters remaining without disappearing on one line. Examples Input 12 3 2 1 1 2 3 2 2 2 1 1 3 12 3 2 1 1 2 3 2 1 3 2 1 3 0 Output 3 12 Input None Output None
instruction
0
12,422
7
24,844
"Correct Solution: ``` def cara(n): kazu=[] ilo=[] a=int(input()) una=a kazu.append(1) ilo.append(a) for n in range(n-1): a=int(input()) if una==a: kazu[-1]+=1 else: kazu.append(1) ilo.append(a) una=a return ilo,kazu def kes(ilo,kazu): k=10000 unanseer=sum(kazu) anseer=sum(kazu) for p in range(len(ilo)): n=0 if kazu[p]>=3 : anseer=kurabe(ilo,kazu,p) n=0 if p+2 < len(ilo) and ilo[p]==ilo[p+2] and kazu[p]+kazu[p+2]>=3 and kazu[p+1]==1: while p+2+n<len(ilo) and p-n>=0 and ilo[p-n]==ilo[p+2+n] and kazu[p-n]+kazu[p+2+n]>=4: n+=1 if n>2500: print(n) unanseer=kensa(ilo,kazu,n,p) if unanseer<anseer: anseer=unanseer if anseer<k: k=anseer return k def kensa(ilo,kazu,n,p): k=0 if n==0: for w in range(len(kazu)): k+=kazu[w] else: for q in range(0,p-n+1): k+=kazu[q] for i in range(p+2+n,len(kazu)): k+=kazu[i] return k def kurabe(ilo,kazu,p): n=0 a=0 b=0 k=0 #δΈŠγ‹γ‚‰γ‚‚γ‚‰γ† if p+1 < len(ilo): kazu[p+1]-=1 kazu[p]+=1 while 0<p-a and len(kazu)>p+1+a and ilo[p+1+a]==ilo[p-1-a] and kazu[p+1+a]+kazu[p-1-a]>=4: a+=1 kazu[p+1]+=1 kazu[p]-=1 #下からもらう if p-1 >= 0: kazu[p-1]-=1 kazu[p]+=1 while len(kazu)>p+1+b and 0<p-b and ilo[p+b+1]==ilo[p-1-b] and kazu[p+b+1]+kazu[p-1-b]>=4: b+=1 kazu[p-1]+=1 kazu[p]-=1 if a<b: n=b else: n=a for q in range(0,p-n): k+=kazu[q] for i in range(p+n,len(kazu)): k+=kazu[i] k-=3 return k while True: n=int(input()) if n==0: break ilo,kazu=cara(n) print(kes(ilo,kazu)) ```
output
1
12,422
7
24,845
Provide a correct Python 3 solution for this coding contest problem. problem There are the following games. N characters are lined up in a vertical row. The color of these characters is red, blue, or yellow, and in the initial state, four or more characters of the same color are not lined up in a row. The player can select a character at a certain position and change it to another color. By this operation, if four or more characters of the same color are lined up in a row, those characters will disappear. When four or more characters of the same color are lined up in a row due to the disappearance of the characters, those characters also disappear, and this chain continues until there are no more places where four or more characters of the same color are lined up in a row. .. The purpose of this game is to reduce the number of characters remaining without disappearing. For example, if the color of the sixth character from the top is changed from yellow to blue in the state at the left end of the figure below, five blue characters will disappear in a row, and finally three characters will remain without disappearing. <image> Given the color sequence of N characters in the initial state, create a program that finds the minimum value M of the number of characters that remain without disappearing when the color of the character is changed in only one place. input The input consists of multiple datasets. Each dataset is given in the following format. The first line consists of only the number of characters N (1 ≀ N ≀ 10000). The following N lines contain one of 1, 2, and 3 integers, and the i + 1st line (1 ≀ i ≀ N) represents the color of the i-th character from the top in the initial state (1). Is red, 2 is blue, and 3 is yellow). When N is 0, it indicates the end of input. The number of datasets does not exceed 5. output For each dataset, output the minimum value M of the number of characters remaining without disappearing on one line. Examples Input 12 3 2 1 1 2 3 2 2 2 1 1 3 12 3 2 1 1 2 3 2 1 3 2 1 3 0 Output 3 12 Input None Output None
instruction
0
12,423
7
24,846
"Correct Solution: ``` def solve(): INF = 100000 while True: n = int(input()) if not n: break lst = [int(input()) for _ in range(n)] def check(x): c = lst[x] l = r = x b = 0 for i in range(r, n): if lst[i] != c: r = i - 1 break else: r = n - 1 for i in range(l, -1, -1): if lst[i] != c: l = i + 1 break else: l = 0 if r - l - b < 3: return n else: b = r - l while l > 0 and r < n - 1: c = lst[l - 1] if c != lst[r + 1]: break else: for i in range(r + 1,n): if lst[i] != c: r = i - 1 break else: r = n - 1 for i in range(l - 1, -1, -1): if lst[i] != c: l = i + 1 break else: l = 0 if r - l - b < 4: break else: b = r - l return n - (b + 1) ans = INF for i in range(n): lst[i] = (lst[i]) % 3 + 1 ans = min(ans, check(i)) lst[i] = (lst[i]) % 3 + 1 ans = min(ans, check(i)) lst[i] = (lst[i]) % 3 + 1 print(ans) solve() ```
output
1
12,423
7
24,847
Provide a correct Python 3 solution for this coding contest problem. problem There are the following games. N characters are lined up in a vertical row. The color of these characters is red, blue, or yellow, and in the initial state, four or more characters of the same color are not lined up in a row. The player can select a character at a certain position and change it to another color. By this operation, if four or more characters of the same color are lined up in a row, those characters will disappear. When four or more characters of the same color are lined up in a row due to the disappearance of the characters, those characters also disappear, and this chain continues until there are no more places where four or more characters of the same color are lined up in a row. .. The purpose of this game is to reduce the number of characters remaining without disappearing. For example, if the color of the sixth character from the top is changed from yellow to blue in the state at the left end of the figure below, five blue characters will disappear in a row, and finally three characters will remain without disappearing. <image> Given the color sequence of N characters in the initial state, create a program that finds the minimum value M of the number of characters that remain without disappearing when the color of the character is changed in only one place. input The input consists of multiple datasets. Each dataset is given in the following format. The first line consists of only the number of characters N (1 ≀ N ≀ 10000). The following N lines contain one of 1, 2, and 3 integers, and the i + 1st line (1 ≀ i ≀ N) represents the color of the i-th character from the top in the initial state (1). Is red, 2 is blue, and 3 is yellow). When N is 0, it indicates the end of input. The number of datasets does not exceed 5. output For each dataset, output the minimum value M of the number of characters remaining without disappearing on one line. Examples Input 12 3 2 1 1 2 3 2 2 2 1 1 3 12 3 2 1 1 2 3 2 1 3 2 1 3 0 Output 3 12 Input None Output None
instruction
0
12,424
7
24,848
"Correct Solution: ``` def solve(): INF = 100000 while True: n = int(input()) if not n: break lst = [int(input()) for _ in range(n)] def check(x): c1 = c2 = lst[x] l = r = x b = 0 for i in range(r, n): if lst[i] != c2: r = i - 1 break else: r = n - 1 for i in range(l, -1, -1): if lst[i] != c1: l = i + 1 break else: l = 0 if r - l - b < 3: return n else: b = r - l while l > 0 and r < n - 1: c1 = lst[l - 1] c2 = lst[r + 1] if c1 != c2: break else: for i in range(r + 1,n): if lst[i] != c2: r = i - 1 break else: r = n - 1 for i in range(l - 1, -1, -1): if lst[i] != c1: l = i + 1 break else: l = 0 if r - l - b < 4: break else: b = r - l return n - (b + 1) ans = INF for i in range(n): lst[i] = (lst[i] + 1) % 3 + 1 ans = min(ans, check(i)) lst[i] = (lst[i] + 1) % 3 + 1 ans = min(ans, check(i)) lst[i] = (lst[i] + 1) % 3 + 1 print(ans) solve() ```
output
1
12,424
7
24,849
Provide a correct Python 3 solution for this coding contest problem. problem There are the following games. N characters are lined up in a vertical row. The color of these characters is red, blue, or yellow, and in the initial state, four or more characters of the same color are not lined up in a row. The player can select a character at a certain position and change it to another color. By this operation, if four or more characters of the same color are lined up in a row, those characters will disappear. When four or more characters of the same color are lined up in a row due to the disappearance of the characters, those characters also disappear, and this chain continues until there are no more places where four or more characters of the same color are lined up in a row. .. The purpose of this game is to reduce the number of characters remaining without disappearing. For example, if the color of the sixth character from the top is changed from yellow to blue in the state at the left end of the figure below, five blue characters will disappear in a row, and finally three characters will remain without disappearing. <image> Given the color sequence of N characters in the initial state, create a program that finds the minimum value M of the number of characters that remain without disappearing when the color of the character is changed in only one place. input The input consists of multiple datasets. Each dataset is given in the following format. The first line consists of only the number of characters N (1 ≀ N ≀ 10000). The following N lines contain one of 1, 2, and 3 integers, and the i + 1st line (1 ≀ i ≀ N) represents the color of the i-th character from the top in the initial state (1). Is red, 2 is blue, and 3 is yellow). When N is 0, it indicates the end of input. The number of datasets does not exceed 5. output For each dataset, output the minimum value M of the number of characters remaining without disappearing on one line. Examples Input 12 3 2 1 1 2 3 2 2 2 1 1 3 12 3 2 1 1 2 3 2 1 3 2 1 3 0 Output 3 12 Input None Output None
instruction
0
12,425
7
24,850
"Correct Solution: ``` INF = 100000 while True: n = int(input()) if not n: break lst = [int(input()) for _ in range(n)] def check(x): c1 = c2 = lst[x] l = r = x b = 0 for i in range(r, n): if lst[i] != c2: r = i - 1 break else: r = n - 1 for i in range(l, -1, -1): if lst[i] != c1: l = i + 1 break else: l = 0 if r - l - b < 3: return n else: b = r - l while l > 0 and r < n - 1: c1 = lst[l - 1] c2 = lst[r + 1] if c1 != c2: break else: for i in range(r + 1,n): if lst[i] != c2: r = i - 1 break else: r = n - 1 for i in range(l - 1, -1, -1): if lst[i] != c1: l = i + 1 break else: l = 0 if r - l - b < 4: break else: b = r - l return n - (b + 1) ans = INF for i in range(n): lst[i] = (lst[i] + 1) % 3 + 1 ans = min(ans, check(i)) lst[i] = (lst[i] + 1) % 3 + 1 ans = min(ans, check(i)) lst[i] = (lst[i] + 1) % 3 + 1 print(ans) ```
output
1
12,425
7
24,851
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. problem There are the following games. N characters are lined up in a vertical row. The color of these characters is red, blue, or yellow, and in the initial state, four or more characters of the same color are not lined up in a row. The player can select a character at a certain position and change it to another color. By this operation, if four or more characters of the same color are lined up in a row, those characters will disappear. When four or more characters of the same color are lined up in a row due to the disappearance of the characters, those characters also disappear, and this chain continues until there are no more places where four or more characters of the same color are lined up in a row. .. The purpose of this game is to reduce the number of characters remaining without disappearing. For example, if the color of the sixth character from the top is changed from yellow to blue in the state at the left end of the figure below, five blue characters will disappear in a row, and finally three characters will remain without disappearing. <image> Given the color sequence of N characters in the initial state, create a program that finds the minimum value M of the number of characters that remain without disappearing when the color of the character is changed in only one place. input The input consists of multiple datasets. Each dataset is given in the following format. The first line consists of only the number of characters N (1 ≀ N ≀ 10000). The following N lines contain one of 1, 2, and 3 integers, and the i + 1st line (1 ≀ i ≀ N) represents the color of the i-th character from the top in the initial state (1). Is red, 2 is blue, and 3 is yellow). When N is 0, it indicates the end of input. The number of datasets does not exceed 5. output For each dataset, output the minimum value M of the number of characters remaining without disappearing on one line. Examples Input 12 3 2 1 1 2 3 2 2 2 1 1 3 12 3 2 1 1 2 3 2 1 3 2 1 3 0 Output 3 12 Input None Output None Submitted Solution: ``` def product4(a): return a[0]*a[1]*a[2]*a[3] def puyo(a): ans = len(a) while True: ans = len(a) for i in range(len(a)-3): if a[i] == a[i+1] == a[i+2] == a[i+3]: hit = a[i] ren = 3 while i+ren+1 < len(a): if a[i+ren+1] != hit: break ren += 1 del a[i:i+ren+1] break else: break return ans while True: n = int(input()) if n==0: break cha = [] for i in range(n): cha.append(int(input())) ans = n for i in range(len(cha)-3): chain = cha.copy() if product4(chain[i:i+4]) in [2,3,8,24,27,54]: hit = chain[i] if hit == chain[i+1]: chain[i+2] = chain[i+3] = hit elif hit == chain[i+2]: chain[i+1] = hit else: chain[i] = chain[i+1] ans = min(ans,puyo(chain)) print(ans) ```
instruction
0
12,426
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24,852
No
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1
12,426
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24,853
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. problem There are the following games. N characters are lined up in a vertical row. The color of these characters is red, blue, or yellow, and in the initial state, four or more characters of the same color are not lined up in a row. The player can select a character at a certain position and change it to another color. By this operation, if four or more characters of the same color are lined up in a row, those characters will disappear. When four or more characters of the same color are lined up in a row due to the disappearance of the characters, those characters also disappear, and this chain continues until there are no more places where four or more characters of the same color are lined up in a row. .. The purpose of this game is to reduce the number of characters remaining without disappearing. For example, if the color of the sixth character from the top is changed from yellow to blue in the state at the left end of the figure below, five blue characters will disappear in a row, and finally three characters will remain without disappearing. <image> Given the color sequence of N characters in the initial state, create a program that finds the minimum value M of the number of characters that remain without disappearing when the color of the character is changed in only one place. input The input consists of multiple datasets. Each dataset is given in the following format. The first line consists of only the number of characters N (1 ≀ N ≀ 10000). The following N lines contain one of 1, 2, and 3 integers, and the i + 1st line (1 ≀ i ≀ N) represents the color of the i-th character from the top in the initial state (1). Is red, 2 is blue, and 3 is yellow). When N is 0, it indicates the end of input. The number of datasets does not exceed 5. output For each dataset, output the minimum value M of the number of characters remaining without disappearing on one line. Examples Input 12 3 2 1 1 2 3 2 2 2 1 1 3 12 3 2 1 1 2 3 2 1 3 2 1 3 0 Output 3 12 Input None Output None Submitted Solution: ``` def product4(a): return a[0]*a[1]*a[2]*a[3] def puyo(a): ans = len(a) for i in range(len(a)-3): if a[i] == a[i+1] == a[i+2] == a[i+3]: hit = a[i] ren = 3 while i+ren+1 < len(a): if a[i+ren+1] != hit: break ren += 1 ans = min(ans, puyo(a[:i]+a[i+ren+1:])) return ans while True: n = int(input()) if n==0: break cha = [] for i in range(n): cha.append(int(input())) ans = n for i in range(len(cha)-3): chain = cha.copy() if product4(chain[i:i+4]) in [2,3,8,24,27,54]: hit = chain[i] if hit == chain[i+1]: chain[i+2] = chain[i+3] = hit elif hit == chain[i+2]: chain[i+1] = hit else: chain[i] = chain[i+1] ans = min(ans,puyo(chain)) print(ans) ```
instruction
0
12,427
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24,854
No
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1
12,427
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24,855
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. problem There are the following games. N characters are lined up in a vertical row. The color of these characters is red, blue, or yellow, and in the initial state, four or more characters of the same color are not lined up in a row. The player can select a character at a certain position and change it to another color. By this operation, if four or more characters of the same color are lined up in a row, those characters will disappear. When four or more characters of the same color are lined up in a row due to the disappearance of the characters, those characters also disappear, and this chain continues until there are no more places where four or more characters of the same color are lined up in a row. .. The purpose of this game is to reduce the number of characters remaining without disappearing. For example, if the color of the sixth character from the top is changed from yellow to blue in the state at the left end of the figure below, five blue characters will disappear in a row, and finally three characters will remain without disappearing. <image> Given the color sequence of N characters in the initial state, create a program that finds the minimum value M of the number of characters that remain without disappearing when the color of the character is changed in only one place. input The input consists of multiple datasets. Each dataset is given in the following format. The first line consists of only the number of characters N (1 ≀ N ≀ 10000). The following N lines contain one of 1, 2, and 3 integers, and the i + 1st line (1 ≀ i ≀ N) represents the color of the i-th character from the top in the initial state (1). Is red, 2 is blue, and 3 is yellow). When N is 0, it indicates the end of input. The number of datasets does not exceed 5. output For each dataset, output the minimum value M of the number of characters remaining without disappearing on one line. Examples Input 12 3 2 1 1 2 3 2 2 2 1 1 3 12 3 2 1 1 2 3 2 1 3 2 1 3 0 Output 3 12 Input None Output None Submitted Solution: ``` while True: n = int(input()) if not n: break puyos = [] pre = int(input()) cnt = 1 for _ in range(n - 1): p = int(input()) if pre != p: puyos.append((pre, cnt)) pre = p cnt = 0 cnt += 1 puyos.append((pre, cnt)) lp = len(puyos) min_puyo = n for i, puyo in enumerate(puyos): color, num = puyo if num > 1: continue tn, j = n, 1 while True: if i - j < 0 or i + j >= lp: break tp_c, tp_n = puyos[i - j] up_c, up_n = puyos[i + j] if tp_c != up_c: break dp = tp_n + up_n + int(j == 1) if dp < 4: break tn -= dp j += 1 if min_puyo > tn: min_puyo = tn print(min_puyo) ```
instruction
0
12,428
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12,428
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Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. problem There are the following games. N characters are lined up in a vertical row. The color of these characters is red, blue, or yellow, and in the initial state, four or more characters of the same color are not lined up in a row. The player can select a character at a certain position and change it to another color. By this operation, if four or more characters of the same color are lined up in a row, those characters will disappear. When four or more characters of the same color are lined up in a row due to the disappearance of the characters, those characters also disappear, and this chain continues until there are no more places where four or more characters of the same color are lined up in a row. .. The purpose of this game is to reduce the number of characters remaining without disappearing. For example, if the color of the sixth character from the top is changed from yellow to blue in the state at the left end of the figure below, five blue characters will disappear in a row, and finally three characters will remain without disappearing. <image> Given the color sequence of N characters in the initial state, create a program that finds the minimum value M of the number of characters that remain without disappearing when the color of the character is changed in only one place. input The input consists of multiple datasets. Each dataset is given in the following format. The first line consists of only the number of characters N (1 ≀ N ≀ 10000). The following N lines contain one of 1, 2, and 3 integers, and the i + 1st line (1 ≀ i ≀ N) represents the color of the i-th character from the top in the initial state (1). Is red, 2 is blue, and 3 is yellow). When N is 0, it indicates the end of input. The number of datasets does not exceed 5. output For each dataset, output the minimum value M of the number of characters remaining without disappearing on one line. Examples Input 12 3 2 1 1 2 3 2 2 2 1 1 3 12 3 2 1 1 2 3 2 1 3 2 1 3 0 Output 3 12 Input None Output None Submitted Solution: ``` import sys r=sys.stdin.readline w={*'123'} for e in iter(r,'0'): n=m=int(e);a=[r()for _ in[0]*n] for i in range(n): for j in w-{a[i]}: b=a[:i]+[j]+a[i+1:] f=1 while f: s=t=f=0;c=b[0] while t<len(b): t=len(b) for x in w-{c}: if x in b[s+1:]:t=min(t,b.index(x,s+1)) if t-s>3:t-=1;f=1;b=b[:s]+b[t+1:];break elif t==len(b):break else:s,c=t,b[t] m=min(m,len(b)) print(m) ```
instruction
0
12,429
7
24,858
No
output
1
12,429
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24,859
Provide tags and a correct Python 3 solution for this coding contest problem. You have a garland consisting of n lamps. Each lamp is colored red, green or blue. The color of the i-th lamp is s_i ('R', 'G' and 'B' β€” colors of lamps in the garland). You have to recolor some lamps in this garland (recoloring a lamp means changing its initial color to another) in such a way that the obtained garland is nice. A garland is called nice if any two lamps of the same color have distance divisible by three between them. I.e. if the obtained garland is t, then for each i, j such that t_i = t_j should be satisfied |i-j|~ mod~ 3 = 0. The value |x| means absolute value of x, the operation x~ mod~ y means remainder of x when divided by y. For example, the following garlands are nice: "RGBRGBRG", "GB", "R", "GRBGRBG", "BRGBRGB". The following garlands are not nice: "RR", "RGBG". Among all ways to recolor the initial garland to make it nice you have to choose one with the minimum number of recolored lamps. If there are multiple optimal solutions, print any of them. Input The first line of the input contains one integer n (1 ≀ n ≀ 2 β‹… 10^5) β€” the number of lamps. The second line of the input contains the string s consisting of n characters 'R', 'G' and 'B' β€” colors of lamps in the garland. Output In the first line of the output print one integer r β€” the minimum number of recolors needed to obtain a nice garland from the given one. In the second line of the output print one string t of length n β€” a nice garland obtained from the initial one with minimum number of recolors. If there are multiple optimal solutions, print any of them. Examples Input 3 BRB Output 1 GRB Input 7 RGBGRBB Output 3 RGBRGBR
instruction
0
12,489
7
24,978
Tags: brute force, greedy, math Correct Solution: ``` n=int(input()) a=input() s=[] for i in a: s.append(i) ans=0 if n==1: print(0) print(a) else: for i in range(1,n-1): if s[i]==s[i-1] and s[i]==s[i+1]: ans+=1 if s[i]=="R": s[i]="G" elif s[i]=="G": s[i]="B" elif s[i]=="B": s[i]="R" elif s[i]==s[i-1]: #print(s) if s[i]=="R" and s[i+1]=="G": s[i]="B" elif s[i]=="R" and s[i+1]=="B": s[i]="G" elif s[i]=="G" and s[i+1]=="R": s[i]="B" elif s[i]=="G" and s[i+1]=="B": s[i]="R" elif s[i]=="B" and s[i+1]=="G": s[i]="R" elif s[i]=="B" and s[i+1]=="R": s[i]="G" ans+=1 #print(s) if s[n-1]==s[n-2]: ans+=1 if s[n-1]=="R": s[n-1]="G" elif s[n-1]=="G": s[n-1]="B" elif s[n-1]=="B": s[n-1]="R" print(ans) for i in s: print(i,end="") print() ```
output
1
12,489
7
24,979
Provide tags and a correct Python 3 solution for this coding contest problem. You have a garland consisting of n lamps. Each lamp is colored red, green or blue. The color of the i-th lamp is s_i ('R', 'G' and 'B' β€” colors of lamps in the garland). You have to recolor some lamps in this garland (recoloring a lamp means changing its initial color to another) in such a way that the obtained garland is nice. A garland is called nice if any two lamps of the same color have distance divisible by three between them. I.e. if the obtained garland is t, then for each i, j such that t_i = t_j should be satisfied |i-j|~ mod~ 3 = 0. The value |x| means absolute value of x, the operation x~ mod~ y means remainder of x when divided by y. For example, the following garlands are nice: "RGBRGBRG", "GB", "R", "GRBGRBG", "BRGBRGB". The following garlands are not nice: "RR", "RGBG". Among all ways to recolor the initial garland to make it nice you have to choose one with the minimum number of recolored lamps. If there are multiple optimal solutions, print any of them. Input The first line of the input contains one integer n (1 ≀ n ≀ 2 β‹… 10^5) β€” the number of lamps. The second line of the input contains the string s consisting of n characters 'R', 'G' and 'B' β€” colors of lamps in the garland. Output In the first line of the output print one integer r β€” the minimum number of recolors needed to obtain a nice garland from the given one. In the second line of the output print one string t of length n β€” a nice garland obtained from the initial one with minimum number of recolors. If there are multiple optimal solutions, print any of them. Examples Input 3 BRB Output 1 GRB Input 7 RGBGRBB Output 3 RGBRGBR
instruction
0
12,490
7
24,980
Tags: brute force, greedy, math Correct Solution: ``` import sys import heapq from math import ceil RI = lambda : [int(x) for x in sys.stdin.readline().split()] ri = lambda : sys.stdin.readline().strip() n = int(ri()) st = ri() dic = {'R':'B', 'B':'G', 'G':'R'} ans = [st[0]] cnt=0 for i in range(1,len(st)): if ans[-1] == st[i]: cnt+=1 if i+1<len(st) and dic[st[i]] != st[i+1]: ans.append(dic[st[i]]) else: ans.append(dic[dic[st[i]]]) else: ans.append(st[i]) print(cnt) for i in ans: print(i,end="") print() ```
output
1
12,490
7
24,981
Provide tags and a correct Python 3 solution for this coding contest problem. You have a garland consisting of n lamps. Each lamp is colored red, green or blue. The color of the i-th lamp is s_i ('R', 'G' and 'B' β€” colors of lamps in the garland). You have to recolor some lamps in this garland (recoloring a lamp means changing its initial color to another) in such a way that the obtained garland is nice. A garland is called nice if any two lamps of the same color have distance divisible by three between them. I.e. if the obtained garland is t, then for each i, j such that t_i = t_j should be satisfied |i-j|~ mod~ 3 = 0. The value |x| means absolute value of x, the operation x~ mod~ y means remainder of x when divided by y. For example, the following garlands are nice: "RGBRGBRG", "GB", "R", "GRBGRBG", "BRGBRGB". The following garlands are not nice: "RR", "RGBG". Among all ways to recolor the initial garland to make it nice you have to choose one with the minimum number of recolored lamps. If there are multiple optimal solutions, print any of them. Input The first line of the input contains one integer n (1 ≀ n ≀ 2 β‹… 10^5) β€” the number of lamps. The second line of the input contains the string s consisting of n characters 'R', 'G' and 'B' β€” colors of lamps in the garland. Output In the first line of the output print one integer r β€” the minimum number of recolors needed to obtain a nice garland from the given one. In the second line of the output print one string t of length n β€” a nice garland obtained from the initial one with minimum number of recolors. If there are multiple optimal solutions, print any of them. Examples Input 3 BRB Output 1 GRB Input 7 RGBGRBB Output 3 RGBRGBR
instruction
0
12,491
7
24,982
Tags: brute force, greedy, math Correct Solution: ``` n = int(input()) arr = list(input().strip()) k = 0 S = ["R", "G", "B"] for i in range(1, n-1): if arr[i]==arr[i-1]: idx = S.index(arr[i]) idx1 = S.index(arr[i+1]) if idx == idx1: idx = (idx+1)%3 arr[i] = S[idx] k += 1 else: tmp = 3 - idx - idx1 arr[i] = S[tmp] k += 1 if n>=2: if arr[n-1]==arr[n-2]: idx = S.index(arr[n-1]) idx = (idx+1)%3 arr[n-1] = S[idx] k += 1 print(k) print("".join(arr)) ```
output
1
12,491
7
24,983
Provide tags and a correct Python 3 solution for this coding contest problem. You have a garland consisting of n lamps. Each lamp is colored red, green or blue. The color of the i-th lamp is s_i ('R', 'G' and 'B' β€” colors of lamps in the garland). You have to recolor some lamps in this garland (recoloring a lamp means changing its initial color to another) in such a way that the obtained garland is nice. A garland is called nice if any two lamps of the same color have distance divisible by three between them. I.e. if the obtained garland is t, then for each i, j such that t_i = t_j should be satisfied |i-j|~ mod~ 3 = 0. The value |x| means absolute value of x, the operation x~ mod~ y means remainder of x when divided by y. For example, the following garlands are nice: "RGBRGBRG", "GB", "R", "GRBGRBG", "BRGBRGB". The following garlands are not nice: "RR", "RGBG". Among all ways to recolor the initial garland to make it nice you have to choose one with the minimum number of recolored lamps. If there are multiple optimal solutions, print any of them. Input The first line of the input contains one integer n (1 ≀ n ≀ 2 β‹… 10^5) β€” the number of lamps. The second line of the input contains the string s consisting of n characters 'R', 'G' and 'B' β€” colors of lamps in the garland. Output In the first line of the output print one integer r β€” the minimum number of recolors needed to obtain a nice garland from the given one. In the second line of the output print one string t of length n β€” a nice garland obtained from the initial one with minimum number of recolors. If there are multiple optimal solutions, print any of them. Examples Input 3 BRB Output 1 GRB Input 7 RGBGRBB Output 3 RGBRGBR
instruction
0
12,492
7
24,984
Tags: brute force, greedy, math Correct Solution: ``` n=int(input()) s=input() lens=len(s) dp=[[0]*5 for _ in range(lens+5)] parent=[[0]*5 for _ in range(lens+5)] R,G,B=1,2,3 sss="0RGB" dp[0][R]=dp[0][G]=dp[0][B]=1 if s[0]=="R": dp[0][R]=0 if s[0]=="G": dp[0][G]=0 if s[0]=="B": dp[0][B]=0 for i in range(1,lens): dp[i][R]=min(dp[i-1][G],dp[i-1][B])+1 if dp[i][R]==dp[i-1][G]+1: parent[i][R]=G else: parent[i][R]=B dp[i][G]=min(dp[i-1][R],dp[i-1][B])+1 if dp[i][G]==dp[i-1][R]+1: parent[i][G]=R else: parent[i][G]=B dp[i][B]=min(dp[i-1][G],dp[i-1][R])+1 if dp[i][B]==dp[i-1][G]+1: parent[i][B]=G else: parent[i][B]=R if s[i]=="R": dp[i][R]-=1 elif s[i]=="G": dp[i][G]-=1 elif s[i]=="B": dp[i][B]-=1 ans=min(dp[lens-1][R],dp[lens-1][G],dp[lens-1][B]) print(min(dp[lens-1][R],dp[lens-1][G],dp[lens-1][B])) ans2=[0]*lens if ans==dp[lens-1][R]: i=lens-1 pp=R while i>=0: ans2[i]=sss[pp] pp=parent[i][pp] i-=1 elif ans==dp[lens-1][G]: i=lens-1 pp=G while i>=0: ans2[i]=sss[pp] pp=parent[i][pp] i-=1 else: i=lens-1 pp=B while i>=0: ans2[i]=sss[pp] pp=parent[i][pp] i-=1 print("".join(ans2)) ```
output
1
12,492
7
24,985
Provide tags and a correct Python 3 solution for this coding contest problem. You have a garland consisting of n lamps. Each lamp is colored red, green or blue. The color of the i-th lamp is s_i ('R', 'G' and 'B' β€” colors of lamps in the garland). You have to recolor some lamps in this garland (recoloring a lamp means changing its initial color to another) in such a way that the obtained garland is nice. A garland is called nice if any two lamps of the same color have distance divisible by three between them. I.e. if the obtained garland is t, then for each i, j such that t_i = t_j should be satisfied |i-j|~ mod~ 3 = 0. The value |x| means absolute value of x, the operation x~ mod~ y means remainder of x when divided by y. For example, the following garlands are nice: "RGBRGBRG", "GB", "R", "GRBGRBG", "BRGBRGB". The following garlands are not nice: "RR", "RGBG". Among all ways to recolor the initial garland to make it nice you have to choose one with the minimum number of recolored lamps. If there are multiple optimal solutions, print any of them. Input The first line of the input contains one integer n (1 ≀ n ≀ 2 β‹… 10^5) β€” the number of lamps. The second line of the input contains the string s consisting of n characters 'R', 'G' and 'B' β€” colors of lamps in the garland. Output In the first line of the output print one integer r β€” the minimum number of recolors needed to obtain a nice garland from the given one. In the second line of the output print one string t of length n β€” a nice garland obtained from the initial one with minimum number of recolors. If there are multiple optimal solutions, print any of them. Examples Input 3 BRB Output 1 GRB Input 7 RGBGRBB Output 3 RGBRGBR
instruction
0
12,493
7
24,986
Tags: brute force, greedy, math Correct Solution: ``` n = int(input()) s = [i for i in input()] if n == 1: print(0) print(''.join(s)) quit() pre = s[0] c = 0 for i in range(1,n-1): if s[i] == pre: if s[i] == 'R' and s[i+1] == 'B':s[i] = 'G';pre = 'G'; elif s[i] == 'R' and s[i+1] == 'G':s[i] = 'B';pre = 'B'; elif s[i] == 'G' and s[i+1] == 'B':s[i] = 'R';pre = 'R'; elif s[i] == 'G' and s[i+1] == 'R':s[i] = 'B';pre = 'B'; elif s[i] == 'B' and s[i+1] == 'G':s[i] = 'R';pre = 'R'; elif s[i] == 'B' and s[i+1] == 'R':s[i] = 'G';pre = 'G'; elif s[i] == 'R' and s[i+1] == 'R':s[i] = 'B';pre = 'B'; elif s[i] == 'B' and s[i+1] == 'B':s[i] = 'R';pre = 'R'; elif s[i] == 'G' and s[i+1] == 'G':s[i] = 'B';pre = 'B'; c += 1 else: pre = s[i] if s[n-1] == pre: c += 1 if s[n-1] == 'R':s[n-1] = 'B' elif s[n-1] == 'B':s[n-1] = 'R' elif s[n-1] == 'G':s[n-1] = 'R' print(c) print(''.join(s)) ```
output
1
12,493
7
24,987
Provide tags and a correct Python 3 solution for this coding contest problem. You have a garland consisting of n lamps. Each lamp is colored red, green or blue. The color of the i-th lamp is s_i ('R', 'G' and 'B' β€” colors of lamps in the garland). You have to recolor some lamps in this garland (recoloring a lamp means changing its initial color to another) in such a way that the obtained garland is nice. A garland is called nice if any two lamps of the same color have distance divisible by three between them. I.e. if the obtained garland is t, then for each i, j such that t_i = t_j should be satisfied |i-j|~ mod~ 3 = 0. The value |x| means absolute value of x, the operation x~ mod~ y means remainder of x when divided by y. For example, the following garlands are nice: "RGBRGBRG", "GB", "R", "GRBGRBG", "BRGBRGB". The following garlands are not nice: "RR", "RGBG". Among all ways to recolor the initial garland to make it nice you have to choose one with the minimum number of recolored lamps. If there are multiple optimal solutions, print any of them. Input The first line of the input contains one integer n (1 ≀ n ≀ 2 β‹… 10^5) β€” the number of lamps. The second line of the input contains the string s consisting of n characters 'R', 'G' and 'B' β€” colors of lamps in the garland. Output In the first line of the output print one integer r β€” the minimum number of recolors needed to obtain a nice garland from the given one. In the second line of the output print one string t of length n β€” a nice garland obtained from the initial one with minimum number of recolors. If there are multiple optimal solutions, print any of them. Examples Input 3 BRB Output 1 GRB Input 7 RGBGRBB Output 3 RGBRGBR
instruction
0
12,494
7
24,988
Tags: brute force, greedy, math Correct Solution: ``` n=int(input()) s=list(input()) m=n//3 m+=1 a=list("GRB") a=a*m a=a[:n] a1=list("GBR") a1=a1*m a1=a1[:n] a2=list("RBG") a2=a2*m a2=a2[:n] a3=list("RGB") a3=a3*m a3=a3[:n] a4=list("BRG") a4=a4*m a4=a4[:n] a5=list("BGR") a5=a5*m a5=a5[:n] ans=[] d=0 for i in range(n): if a[i]!=s[i]: d+=1 ans.append([d,0]) d=0 for i in range(n): if a1[i]!=s[i]: d+=1 ans.append([d,1]) d=0 for i in range(n): if a2[i]!=s[i]: d+=1 ans.append([d,2]) d=0 for i in range(n): if a3[i]!=s[i]: d+=1 ans.append([d,3]) d=0 for i in range(n): if a4[i]!=s[i]: d+=1 ans.append([d,4]) d=0 for i in range(n): if a5[i]!=s[i]: d+=1 ans.append([d,5]) ans.sort() if ans[0][1]==0: print(ans[0][0]) print(''.join(a)) elif ans[0][1]==1: print(ans[0][0]) print(''.join(a1)) elif ans[0][1]==2: print(ans[0][0]) print(''.join(a2)) elif ans[0][1]==3: print(ans[0][0]) print(''.join(a3)) elif ans[0][1]==4: print(ans[0][0]) print(''.join(a4)) elif ans[0][1]==5: print(ans[0][0]) print(''.join(a5)) ```
output
1
12,494
7
24,989
Provide tags and a correct Python 3 solution for this coding contest problem. You have a garland consisting of n lamps. Each lamp is colored red, green or blue. The color of the i-th lamp is s_i ('R', 'G' and 'B' β€” colors of lamps in the garland). You have to recolor some lamps in this garland (recoloring a lamp means changing its initial color to another) in such a way that the obtained garland is nice. A garland is called nice if any two lamps of the same color have distance divisible by three between them. I.e. if the obtained garland is t, then for each i, j such that t_i = t_j should be satisfied |i-j|~ mod~ 3 = 0. The value |x| means absolute value of x, the operation x~ mod~ y means remainder of x when divided by y. For example, the following garlands are nice: "RGBRGBRG", "GB", "R", "GRBGRBG", "BRGBRGB". The following garlands are not nice: "RR", "RGBG". Among all ways to recolor the initial garland to make it nice you have to choose one with the minimum number of recolored lamps. If there are multiple optimal solutions, print any of them. Input The first line of the input contains one integer n (1 ≀ n ≀ 2 β‹… 10^5) β€” the number of lamps. The second line of the input contains the string s consisting of n characters 'R', 'G' and 'B' β€” colors of lamps in the garland. Output In the first line of the output print one integer r β€” the minimum number of recolors needed to obtain a nice garland from the given one. In the second line of the output print one string t of length n β€” a nice garland obtained from the initial one with minimum number of recolors. If there are multiple optimal solutions, print any of them. Examples Input 3 BRB Output 1 GRB Input 7 RGBGRBB Output 3 RGBRGBR
instruction
0
12,495
7
24,990
Tags: brute force, greedy, math Correct Solution: ``` from math import* n=int(input()) s=list(map(str,input())) t=0 ot=0 while t<len(s)-2: if s[t]==s[t+1]: if (s[t]=='R' or s[t+2]=='R') and (s[t]=='B' or s[t+2]=='B'): s[t+1]='G' ot+=1 elif (s[t]=='R' or s[t+2]=='R') and (s[t]=='G' or s[t+2]=='G'): s[t+1]='B' ot+=1 elif (s[t]=='G' or s[t+2]=='G') and (s[t]=='B' or s[t+2]=='B'): s[t+1]='R' ot+=1 else: if s[t]=='G' or s[t]=='B': s[t+1]='R' ot+=1 else: s[t+1]='G' ot+=1 t+=1 if len(s)>=2: if s[t]==s[t+1]: if s[t]=='G' or s[t]=='B': s[t+1]='R' ot+=1 else: s[t+1]='G' ot+=1 print(ot) print(''.join(s)) ```
output
1
12,495
7
24,991
Provide tags and a correct Python 3 solution for this coding contest problem. You have a garland consisting of n lamps. Each lamp is colored red, green or blue. The color of the i-th lamp is s_i ('R', 'G' and 'B' β€” colors of lamps in the garland). You have to recolor some lamps in this garland (recoloring a lamp means changing its initial color to another) in such a way that the obtained garland is nice. A garland is called nice if any two lamps of the same color have distance divisible by three between them. I.e. if the obtained garland is t, then for each i, j such that t_i = t_j should be satisfied |i-j|~ mod~ 3 = 0. The value |x| means absolute value of x, the operation x~ mod~ y means remainder of x when divided by y. For example, the following garlands are nice: "RGBRGBRG", "GB", "R", "GRBGRBG", "BRGBRGB". The following garlands are not nice: "RR", "RGBG". Among all ways to recolor the initial garland to make it nice you have to choose one with the minimum number of recolored lamps. If there are multiple optimal solutions, print any of them. Input The first line of the input contains one integer n (1 ≀ n ≀ 2 β‹… 10^5) β€” the number of lamps. The second line of the input contains the string s consisting of n characters 'R', 'G' and 'B' β€” colors of lamps in the garland. Output In the first line of the output print one integer r β€” the minimum number of recolors needed to obtain a nice garland from the given one. In the second line of the output print one string t of length n β€” a nice garland obtained from the initial one with minimum number of recolors. If there are multiple optimal solutions, print any of them. Examples Input 3 BRB Output 1 GRB Input 7 RGBGRBB Output 3 RGBRGBR
instruction
0
12,496
7
24,992
Tags: brute force, greedy, math Correct Solution: ``` q = int(input()) Grlnda = list(input()) ex1 = ['R','G','B']; c1=0; line1="" ex2= ['G','R','B']; c2=0; line2="" ex3= ['G','B','R']; c3=0; line3="" ex4= ['R','B','G']; c4=0; line4="" ex5= ['B','G','R']; c5=0; line5="" ex6= ['B','R','G']; c6=0; line6="" for i in range(q): line1 = line1 + ex1[i % 3] line2 = line2 + ex2[i % 3] line3 = line3 + ex3[i % 3] line4 = line4 + ex4[i % 3] line5 = line5 + ex5[i % 3] line6 = line6 + ex6[i % 3] if Grlnda[i]!=ex1[i%3]: c1 = c1+1 if Grlnda[i]!=ex2[i%3]: c2 = c2+1 if Grlnda[i]!=ex3[i%3]: c3 = c3+1 if Grlnda[i]!=ex4[i%3]: c4 = c4+1 if Grlnda[i]!=ex5[i%3]: c5 = c5+1 if Grlnda[i]!=ex6[i%3]: c6 = c6+1 numlist = [c1,c2,c3,c4,c5,c6] strlist = [line1,line2,line3,line4,line5,line6] print(min(numlist)) print(strlist[numlist.index(min(numlist))]) ```
output
1
12,496
7
24,993
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. You have a garland consisting of n lamps. Each lamp is colored red, green or blue. The color of the i-th lamp is s_i ('R', 'G' and 'B' β€” colors of lamps in the garland). You have to recolor some lamps in this garland (recoloring a lamp means changing its initial color to another) in such a way that the obtained garland is nice. A garland is called nice if any two lamps of the same color have distance divisible by three between them. I.e. if the obtained garland is t, then for each i, j such that t_i = t_j should be satisfied |i-j|~ mod~ 3 = 0. The value |x| means absolute value of x, the operation x~ mod~ y means remainder of x when divided by y. For example, the following garlands are nice: "RGBRGBRG", "GB", "R", "GRBGRBG", "BRGBRGB". The following garlands are not nice: "RR", "RGBG". Among all ways to recolor the initial garland to make it nice you have to choose one with the minimum number of recolored lamps. If there are multiple optimal solutions, print any of them. Input The first line of the input contains one integer n (1 ≀ n ≀ 2 β‹… 10^5) β€” the number of lamps. The second line of the input contains the string s consisting of n characters 'R', 'G' and 'B' β€” colors of lamps in the garland. Output In the first line of the output print one integer r β€” the minimum number of recolors needed to obtain a nice garland from the given one. In the second line of the output print one string t of length n β€” a nice garland obtained from the initial one with minimum number of recolors. If there are multiple optimal solutions, print any of them. Examples Input 3 BRB Output 1 GRB Input 7 RGBGRBB Output 3 RGBRGBR Submitted Solution: ``` n=int(input()) l=list(input()) i=1 c=0 while i<(n-1): if l[i]==l[i-1]: c+=1 if l[i]=='R': if l[i+1]=='G': l[i]='B' else: l[i]='G' elif l[i]=='G': if l[i+1]=='R': l[i]='B' else: l[i]='R' elif l[i]=='B': if l[i+1]=='R': l[i]='G' else: l[i]='R' i+=1 if n!=1: if l[n-1]==l[n-2]: c+=1 if l[n-1]=='R': l[n-1]='G' else: l[n-1]='R' print(c) print(''.join(l)) ```
instruction
0
12,497
7
24,994
Yes
output
1
12,497
7
24,995
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. You have a garland consisting of n lamps. Each lamp is colored red, green or blue. The color of the i-th lamp is s_i ('R', 'G' and 'B' β€” colors of lamps in the garland). You have to recolor some lamps in this garland (recoloring a lamp means changing its initial color to another) in such a way that the obtained garland is nice. A garland is called nice if any two lamps of the same color have distance divisible by three between them. I.e. if the obtained garland is t, then for each i, j such that t_i = t_j should be satisfied |i-j|~ mod~ 3 = 0. The value |x| means absolute value of x, the operation x~ mod~ y means remainder of x when divided by y. For example, the following garlands are nice: "RGBRGBRG", "GB", "R", "GRBGRBG", "BRGBRGB". The following garlands are not nice: "RR", "RGBG". Among all ways to recolor the initial garland to make it nice you have to choose one with the minimum number of recolored lamps. If there are multiple optimal solutions, print any of them. Input The first line of the input contains one integer n (1 ≀ n ≀ 2 β‹… 10^5) β€” the number of lamps. The second line of the input contains the string s consisting of n characters 'R', 'G' and 'B' β€” colors of lamps in the garland. Output In the first line of the output print one integer r β€” the minimum number of recolors needed to obtain a nice garland from the given one. In the second line of the output print one string t of length n β€” a nice garland obtained from the initial one with minimum number of recolors. If there are multiple optimal solutions, print any of them. Examples Input 3 BRB Output 1 GRB Input 7 RGBGRBB Output 3 RGBRGBR Submitted Solution: ``` n = int(input()) s = list(str(input())) changesMade = 0 i=0 while i < n-1: if s[i] == s[i+1]: j = i + 0 while j < n-1 and s[j] == s[j+1]: if (j-i) % 2 == 0: k = j+1 if k >= n-2: if s[k-1] != "B": s[k] = "B" else: s[k] = "G" elif s[k-1] != "G" and s[k+1] != "G": s[k] = "G" elif s[k-1] != "B" and s[k+1] != "B": s[k] = "B" elif s[k-1] != "R" and s[k+1] != "R": s[k] = "R" else: print("ERROR") changesMade+=1 j+=1 i+=1 print(changesMade) print("".join(s)) ```
instruction
0
12,498
7
24,996
Yes
output
1
12,498
7
24,997
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. You have a garland consisting of n lamps. Each lamp is colored red, green or blue. The color of the i-th lamp is s_i ('R', 'G' and 'B' β€” colors of lamps in the garland). You have to recolor some lamps in this garland (recoloring a lamp means changing its initial color to another) in such a way that the obtained garland is nice. A garland is called nice if any two lamps of the same color have distance divisible by three between them. I.e. if the obtained garland is t, then for each i, j such that t_i = t_j should be satisfied |i-j|~ mod~ 3 = 0. The value |x| means absolute value of x, the operation x~ mod~ y means remainder of x when divided by y. For example, the following garlands are nice: "RGBRGBRG", "GB", "R", "GRBGRBG", "BRGBRGB". The following garlands are not nice: "RR", "RGBG". Among all ways to recolor the initial garland to make it nice you have to choose one with the minimum number of recolored lamps. If there are multiple optimal solutions, print any of them. Input The first line of the input contains one integer n (1 ≀ n ≀ 2 β‹… 10^5) β€” the number of lamps. The second line of the input contains the string s consisting of n characters 'R', 'G' and 'B' β€” colors of lamps in the garland. Output In the first line of the output print one integer r β€” the minimum number of recolors needed to obtain a nice garland from the given one. In the second line of the output print one string t of length n β€” a nice garland obtained from the initial one with minimum number of recolors. If there are multiple optimal solutions, print any of them. Examples Input 3 BRB Output 1 GRB Input 7 RGBGRBB Output 3 RGBRGBR Submitted Solution: ``` def permutations(string): """Create all permutations of a string with non-repeating characters """ permutation_list = [] if len(string) == 1: return [string] else: for char in string: [permutation_list.append(char + a) for a in permutations(string.replace(char, ""))] return permutation_list def comp(l1): count = 0 for i in range(n): if(l1[i] != color[i]): count+=1 return count n = int(input()) tab = [0] * n color = list(input()) #print(color) tab1 = ['R'] * n tab2 = ['B'] * n tab3 = ['G'] * n for i in range(n): tab[i] = i % 3 #print(tab) perm = permutations('RGB') tabPoss = [ ['R' for j in range(n)] for i in range(6)] #print(tabPoss) count = 0 for e in perm: for i in range(n): tabPoss[count][i] = e[i%3] count+=1 #print(tabPoss) res = n + 1 top = -1 for i in range(6): data = comp(tabPoss[i]) if(res > data): res = data top = i print(res) print(''.join(tabPoss[top])) ```
instruction
0
12,499
7
24,998
Yes
output
1
12,499
7
24,999
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. You have a garland consisting of n lamps. Each lamp is colored red, green or blue. The color of the i-th lamp is s_i ('R', 'G' and 'B' β€” colors of lamps in the garland). You have to recolor some lamps in this garland (recoloring a lamp means changing its initial color to another) in such a way that the obtained garland is nice. A garland is called nice if any two lamps of the same color have distance divisible by three between them. I.e. if the obtained garland is t, then for each i, j such that t_i = t_j should be satisfied |i-j|~ mod~ 3 = 0. The value |x| means absolute value of x, the operation x~ mod~ y means remainder of x when divided by y. For example, the following garlands are nice: "RGBRGBRG", "GB", "R", "GRBGRBG", "BRGBRGB". The following garlands are not nice: "RR", "RGBG". Among all ways to recolor the initial garland to make it nice you have to choose one with the minimum number of recolored lamps. If there are multiple optimal solutions, print any of them. Input The first line of the input contains one integer n (1 ≀ n ≀ 2 β‹… 10^5) β€” the number of lamps. The second line of the input contains the string s consisting of n characters 'R', 'G' and 'B' β€” colors of lamps in the garland. Output In the first line of the output print one integer r β€” the minimum number of recolors needed to obtain a nice garland from the given one. In the second line of the output print one string t of length n β€” a nice garland obtained from the initial one with minimum number of recolors. If there are multiple optimal solutions, print any of them. Examples Input 3 BRB Output 1 GRB Input 7 RGBGRBB Output 3 RGBRGBR Submitted Solution: ``` input() s = input() orders = ['RGB', 'RBG', 'GRB', 'GBR', 'BRG', 'BGR'] garlands = [''.join(o[i % 3] for i in range(len(s))) for o in orders] with_changes = [(g, sum(int(c1 != c2) for c1, c2 in zip(g, s))) for g in garlands] g, c = min(with_changes, key=lambda t: t[1]) print(c) print(g) ```
instruction
0
12,500
7
25,000
Yes
output
1
12,500
7
25,001
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. You have a garland consisting of n lamps. Each lamp is colored red, green or blue. The color of the i-th lamp is s_i ('R', 'G' and 'B' β€” colors of lamps in the garland). You have to recolor some lamps in this garland (recoloring a lamp means changing its initial color to another) in such a way that the obtained garland is nice. A garland is called nice if any two lamps of the same color have distance divisible by three between them. I.e. if the obtained garland is t, then for each i, j such that t_i = t_j should be satisfied |i-j|~ mod~ 3 = 0. The value |x| means absolute value of x, the operation x~ mod~ y means remainder of x when divided by y. For example, the following garlands are nice: "RGBRGBRG", "GB", "R", "GRBGRBG", "BRGBRGB". The following garlands are not nice: "RR", "RGBG". Among all ways to recolor the initial garland to make it nice you have to choose one with the minimum number of recolored lamps. If there are multiple optimal solutions, print any of them. Input The first line of the input contains one integer n (1 ≀ n ≀ 2 β‹… 10^5) β€” the number of lamps. The second line of the input contains the string s consisting of n characters 'R', 'G' and 'B' β€” colors of lamps in the garland. Output In the first line of the output print one integer r β€” the minimum number of recolors needed to obtain a nice garland from the given one. In the second line of the output print one string t of length n β€” a nice garland obtained from the initial one with minimum number of recolors. If there are multiple optimal solutions, print any of them. Examples Input 3 BRB Output 1 GRB Input 7 RGBGRBB Output 3 RGBRGBR Submitted Solution: ``` n=int(input()) s1=input() s=list(s1) c=0 a=[0]*n d=['R','G','B'] if(len(s)==3): for i in range(len(s)-2): if(s[i]==s[i+2]): c=1 if(s[i]=='B' and s[i+1]=='R'): s[i+2]='G' if(s[i]=='B' and s[i+1]=='G'): s[i+2]='R' if(s[i]=='G' and s[i+1]=='R'): s[i+2]='B' if(s[i]=='G' and s[i+1]=='B'): s[i+2]='R' if(s[i]=='R' and s[i+1]=='G'): s[i+2]='B' if(s[i]=='R' and s[i+1]=='B'): s[i+2]='G' elif(s[i+1]==s[i+2]): c=1 if(s[i]=='B' and s[i+1]=='R'): s[i+2]='G' if(s[i]=='B' and s[i+1]=='G'): s[i+2]='R' if(s[i]=='G' and s[i+1]=='R'): s[i+2]='B' if(s[i]=='G' and s[i+1]=='B'): s[i+2]='R' if(s[i]=='R' and s[i+1]=='G'): s[i]='B' if(s[i+2]=='R' and s[i+1]=='B'): s[i]='G' elif(s[i]==s[i+1]==s[i+2]): c=2 if(s[i]=='B'): s[i+1]='G' s[i+2]='R' if(s[i]=='G'): s[i+1]='B' s[i+2]='R' if(s[i]=='R'): s[i+1]='G' s[i+2]='B' if(len(s)==2): if(s[0]==s[1]): c=1 if(s[1]=='R'): s[0]='G' if(s[1]=='B'): s[0]='G' if(s[1]=='G'): s[0]='B' if(len(s)==1): c=0 else: for i in range(0,len(s)-3,1): if(s[i]!=s[i+3]): s[i+3]=s[i] c+=1 print(c) for i in s: print(i, end="") ```
instruction
0
12,501
7
25,002
No
output
1
12,501
7
25,003
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. You have a garland consisting of n lamps. Each lamp is colored red, green or blue. The color of the i-th lamp is s_i ('R', 'G' and 'B' β€” colors of lamps in the garland). You have to recolor some lamps in this garland (recoloring a lamp means changing its initial color to another) in such a way that the obtained garland is nice. A garland is called nice if any two lamps of the same color have distance divisible by three between them. I.e. if the obtained garland is t, then for each i, j such that t_i = t_j should be satisfied |i-j|~ mod~ 3 = 0. The value |x| means absolute value of x, the operation x~ mod~ y means remainder of x when divided by y. For example, the following garlands are nice: "RGBRGBRG", "GB", "R", "GRBGRBG", "BRGBRGB". The following garlands are not nice: "RR", "RGBG". Among all ways to recolor the initial garland to make it nice you have to choose one with the minimum number of recolored lamps. If there are multiple optimal solutions, print any of them. Input The first line of the input contains one integer n (1 ≀ n ≀ 2 β‹… 10^5) β€” the number of lamps. The second line of the input contains the string s consisting of n characters 'R', 'G' and 'B' β€” colors of lamps in the garland. Output In the first line of the output print one integer r β€” the minimum number of recolors needed to obtain a nice garland from the given one. In the second line of the output print one string t of length n β€” a nice garland obtained from the initial one with minimum number of recolors. If there are multiple optimal solutions, print any of them. Examples Input 3 BRB Output 1 GRB Input 7 RGBGRBB Output 3 RGBRGBR Submitted Solution: ``` n=int(input()) sa=[0,0,0] la=[0,0,0] ka=[0,0,0] p=input() x=0 y=1 z=2 while x<n: if p[x]=="B": sa[0]+=1 elif p[x]=="G": sa[1]+=1 else: sa[2]+=1 if y<n: if p[y]=="B": la[0]+=1 elif p[y]=="G": la[1]+=1 else: la[2]+=1 if z<n: if p[z]=="B": ka[0]+=1 elif p[z]=="G": ka[1]+=1 else: ka[2]+=1 x+=3 y+=3 z+=3 aa=sa.index(max(sa)) ba=la.index(max(la)) ca=ka.index(max(ka)) fam=[0,0,0] if aa==0: fam[0]="B" if aa==1: fam[0]="G" if aa==2: fam[0]="R" if ba==0: fam[1]="B" if ba==1: fam[1]="G" if ba==2: fam[1]="R" if ca==0: fam[2]="B" if ca==1: fam[2]="G" if ca==2: fam[2]="R" jam=fam*n jam=jam[0:n] s=0 for x in range(n): if p[x]==jam[x]: pass else: s+=1 if n==2: if p[0]!=p[1]: print(0) print(p) else: print(1) if p[0]!="B": print(p[0]+"B") else: print(p[0]+"R") elif n==3: if len(set(p))==n: print(0) print(p) elif len(set(p))==n-1: p=list(p) if p[0]==p[1]: if p[2]=="R": p[0]="B" p[1]="G" elif p[2]=="G": p[0]="B" p[1]="R" else: p[0]="G" p[1]="R" if p[0]==p[2]: if p[1]=="R": p[0]="B" p[2]="G" elif p[1]=="G": p[0]="B" p[2]="R" else: p[0]="G" p[2]="R" if p[1]==p[2]: if p[0]=="R": p[2]="B" p[1]="G" elif p[0]=="G": p[2]="B" p[1]="R" else: p[2]="G" p[1]="R" print(1) print("".join(p)) else: p=list(p) print(2) if p[0]=="B": print(p[0]+"G"+"R") elif p[0]=="G": print(p[0]+"B"+"R") else: print(p[0]+"B"+"G") elif n==1: print(0) print(p) else: print(s) print("".join(jam)) ```
instruction
0
12,502
7
25,004
No
output
1
12,502
7
25,005
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. You have a garland consisting of n lamps. Each lamp is colored red, green or blue. The color of the i-th lamp is s_i ('R', 'G' and 'B' β€” colors of lamps in the garland). You have to recolor some lamps in this garland (recoloring a lamp means changing its initial color to another) in such a way that the obtained garland is nice. A garland is called nice if any two lamps of the same color have distance divisible by three between them. I.e. if the obtained garland is t, then for each i, j such that t_i = t_j should be satisfied |i-j|~ mod~ 3 = 0. The value |x| means absolute value of x, the operation x~ mod~ y means remainder of x when divided by y. For example, the following garlands are nice: "RGBRGBRG", "GB", "R", "GRBGRBG", "BRGBRGB". The following garlands are not nice: "RR", "RGBG". Among all ways to recolor the initial garland to make it nice you have to choose one with the minimum number of recolored lamps. If there are multiple optimal solutions, print any of them. Input The first line of the input contains one integer n (1 ≀ n ≀ 2 β‹… 10^5) β€” the number of lamps. The second line of the input contains the string s consisting of n characters 'R', 'G' and 'B' β€” colors of lamps in the garland. Output In the first line of the output print one integer r β€” the minimum number of recolors needed to obtain a nice garland from the given one. In the second line of the output print one string t of length n β€” a nice garland obtained from the initial one with minimum number of recolors. If there are multiple optimal solutions, print any of them. Examples Input 3 BRB Output 1 GRB Input 7 RGBGRBB Output 3 RGBRGBR Submitted Solution: ``` n = int(input()) s = input() b = ['R','G','B'] count = 0 k = len(s) if len(s) == 1: print(count) print(s) else: for i in range(k-3): if s[i] == s[i+1]: for j in b: if j != s[i] and j != s[i+2]: s = s[:i+1]+j+s[i+2:] count+=1 break if s[k-2] == s[k-1]: for j in b: if j != s[k-2]: s = s[:k-2]+j+s[k-1:] count+=1 break print(count) print(s) ```
instruction
0
12,503
7
25,006
No
output
1
12,503
7
25,007
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. You have a garland consisting of n lamps. Each lamp is colored red, green or blue. The color of the i-th lamp is s_i ('R', 'G' and 'B' β€” colors of lamps in the garland). You have to recolor some lamps in this garland (recoloring a lamp means changing its initial color to another) in such a way that the obtained garland is nice. A garland is called nice if any two lamps of the same color have distance divisible by three between them. I.e. if the obtained garland is t, then for each i, j such that t_i = t_j should be satisfied |i-j|~ mod~ 3 = 0. The value |x| means absolute value of x, the operation x~ mod~ y means remainder of x when divided by y. For example, the following garlands are nice: "RGBRGBRG", "GB", "R", "GRBGRBG", "BRGBRGB". The following garlands are not nice: "RR", "RGBG". Among all ways to recolor the initial garland to make it nice you have to choose one with the minimum number of recolored lamps. If there are multiple optimal solutions, print any of them. Input The first line of the input contains one integer n (1 ≀ n ≀ 2 β‹… 10^5) β€” the number of lamps. The second line of the input contains the string s consisting of n characters 'R', 'G' and 'B' β€” colors of lamps in the garland. Output In the first line of the output print one integer r β€” the minimum number of recolors needed to obtain a nice garland from the given one. In the second line of the output print one string t of length n β€” a nice garland obtained from the initial one with minimum number of recolors. If there are multiple optimal solutions, print any of them. Examples Input 3 BRB Output 1 GRB Input 7 RGBGRBB Output 3 RGBRGBR Submitted Solution: ``` n = int(input()) s = input() if n == 1: print(n) print(s) else: A = [[0]*3 for i in range(n)] Aid = [[0]*3 for i in range(n)] M = {'R':0,'G':1,'B':2} M_ = {0:'R',1:'G',2:'B'} v = M[s[0]] A[0][(v+1)%3]= 1 A[0][(v+2)%3] = 1 def whichmax(A,id): if A[id][0] <= A[id][1] and A[id][0] <= A[id][2]: return 0 if A[id][1] <= A[id][0] and A[id][1] <= A[id][2]: return 1 return 2 def min_and_id(a,ida,b,idb): if a<b: return a,ida else: return b,idb for i in range(1,n): v = M[s[i]] vo0 = A[i-1][v] vo1 = A[i-1][(v+1)%3] vo2 = A[i-1][(v+2)%3] val, vid = min_and_id(vo1,(v+1)%3,vo2,(v+2)%3) #self A[i][v] = val Aid[i][v] = vid val2,vid2 = min_and_id(vo0,v,vo2,(v+2)%3) #N-1 A[i][(v+1)%3] = val2 + 1 Aid[i][(v + 1) % 3] = vid2 val3, vid3 = min_and_id(vo0, v, vo1, (v + 1) % 3) #N-2 A[i][(v + 2) % 3] = val3 + 1 Aid[i][(v + 2) % 3] = vid3 ans = "" i = n-1 midx = whichmax(A,n-1) ans += M_[midx] pid = Aid[n-1][midx] while i>0: ans += M_[pid] pid = Aid[i][pid] i-=1 print(A[n-1][midx]) print(ans[::-1]) ```
instruction
0
12,504
7
25,008
No
output
1
12,504
7
25,009
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Nanami is an expert at playing games. This day, Nanami's good friend Hajime invited her to watch a game of baseball. Unwilling as she was, she followed him to the stadium. But Nanami had no interest in the game, so she looked around to see if there was something that might interest her. That's when she saw the digital board at one end of the stadium. The digital board is n pixels in height and m pixels in width, every pixel is either light or dark. The pixels are described by its coordinate. The j-th pixel of the i-th line is pixel (i, j). The board displays messages by switching a combination of pixels to light, and the rest to dark. Nanami notices that the state of the pixels on the board changes from time to time. At certain times, certain pixels on the board may switch from light to dark, or from dark to light. Nanami wonders, what is the area of the biggest light block such that a specific pixel is on its side. A light block is a sub-rectangle of the board, in which all pixels are light. Pixel (i, j) belongs to a side of sub-rectangle with (x1, y1) and (x2, y2) as its upper-left and lower-right vertex if and only if it satisfies the logical condition: ((i = x1 or i = x2) and (y1 ≀ j ≀ y2)) or ((j = y1 or j = y2) and (x1 ≀ i ≀ x2)). Nanami has all the history of changing pixels, also she has some questions of the described type, can you answer them? Input The first line contains three space-separated integers n, m and q (1 ≀ n, m, q ≀ 1000) β€” the height and width of the digital board, and the number of operations. Then follow n lines, each line containing m space-separated integers. The j-th integer of the i-th line is ai, j β€” the initial state of pixel (i, j). * If ai, j = 0, pixel (i, j) is initially dark. * If ai, j = 1, pixel (i, j) is initially light. Then follow q lines, each line containing three space-separated integers op, x, and y (1 ≀ op ≀ 2; 1 ≀ x ≀ n; 1 ≀ y ≀ m), describing an operation. * If op = 1, the pixel at (x, y) changes its state (from light to dark or from dark to light). * If op = 2, Nanami queries the biggest light block with pixel (x, y) on its side. Output For each query, print a single line containing one integer β€” the answer to Nanami's query. Examples Input 3 4 5 0 1 1 0 1 0 0 1 0 1 1 0 2 2 2 2 1 2 1 2 2 1 2 3 2 2 2 Output 0 2 6 Input 3 3 4 1 1 1 1 1 1 1 1 1 2 2 2 1 2 2 2 1 1 2 2 1 Output 6 3 3 Note Consider the first sample. The first query specifies pixel (2, 2), which is dark itself, so there are no valid light blocks, thus the answer is 0. The second query specifies pixel (1, 2). The biggest light block is the block with (1, 2) as its upper-left vertex and (1, 3) as its lower-right vertex. The last query specifies pixel (2, 2), which became light in the third operation. The biggest light block is the block with (1, 2) as its upper-left vertex and (3, 3) as its lower-right vertex. Submitted Solution: ``` n, m, q = [int(c) for c in input().split()] field = [] for i in range(n): field.append([int(c) for c in input().split()]) def max_dim(x, y, dx, dy): ix = x iy = y count = 0 while (0 <= ix < m) and (0 <= iy < n) and (field[iy][ix] == 1): ix += dx iy += dy count += 1 return count def find_square(arr, value): f = 0 while arr[f] < value: f += 1 l = len(arr) - 1 while arr[l] < value: l -= 1 return (l - f + 1) * value def count_one(x, y, dx, dy): if dx == 1: x_ = x y_ = y max_y_ = n else: x_ = y y_ = x max_y_ = m counter = [] for i in range(max_y_): if dx == 1: counter.append(max_dim(x_, i, dx, dy)) else: counter.append(max_dim(i, x_, dx, dy)) for i in range(y_ - 1, -1, -1): if counter[i + 1] < counter[i]: counter[i] = counter[i + 1] for i in range(y_ + 1, max_y_): if counter[i - 1] < counter[i]: counter[i] = counter[i - 1] max = 0 for i in range(max_y_): t = find_square(counter, counter[i]) if t > max: max = t return max for i in range(q): order, x, y = [int(c) for c in input().split()] x -= 1 y -= 1 if order == 1: field[x][y] = (field[x][y] + 1) % 2 if order == 2: print(max(count_one(y, x, 1, 0),count_one(y, x, -1, 0),count_one(y, x, 0, 1),count_one(y, x, 0, -1))) # print(field) # print(find_square([2,2,3,4,3,1,1,1,1,0], 4)) ```
instruction
0
12,825
7
25,650
No
output
1
12,825
7
25,651
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Nanami is an expert at playing games. This day, Nanami's good friend Hajime invited her to watch a game of baseball. Unwilling as she was, she followed him to the stadium. But Nanami had no interest in the game, so she looked around to see if there was something that might interest her. That's when she saw the digital board at one end of the stadium. The digital board is n pixels in height and m pixels in width, every pixel is either light or dark. The pixels are described by its coordinate. The j-th pixel of the i-th line is pixel (i, j). The board displays messages by switching a combination of pixels to light, and the rest to dark. Nanami notices that the state of the pixels on the board changes from time to time. At certain times, certain pixels on the board may switch from light to dark, or from dark to light. Nanami wonders, what is the area of the biggest light block such that a specific pixel is on its side. A light block is a sub-rectangle of the board, in which all pixels are light. Pixel (i, j) belongs to a side of sub-rectangle with (x1, y1) and (x2, y2) as its upper-left and lower-right vertex if and only if it satisfies the logical condition: ((i = x1 or i = x2) and (y1 ≀ j ≀ y2)) or ((j = y1 or j = y2) and (x1 ≀ i ≀ x2)). Nanami has all the history of changing pixels, also she has some questions of the described type, can you answer them? Input The first line contains three space-separated integers n, m and q (1 ≀ n, m, q ≀ 1000) β€” the height and width of the digital board, and the number of operations. Then follow n lines, each line containing m space-separated integers. The j-th integer of the i-th line is ai, j β€” the initial state of pixel (i, j). * If ai, j = 0, pixel (i, j) is initially dark. * If ai, j = 1, pixel (i, j) is initially light. Then follow q lines, each line containing three space-separated integers op, x, and y (1 ≀ op ≀ 2; 1 ≀ x ≀ n; 1 ≀ y ≀ m), describing an operation. * If op = 1, the pixel at (x, y) changes its state (from light to dark or from dark to light). * If op = 2, Nanami queries the biggest light block with pixel (x, y) on its side. Output For each query, print a single line containing one integer β€” the answer to Nanami's query. Examples Input 3 4 5 0 1 1 0 1 0 0 1 0 1 1 0 2 2 2 2 1 2 1 2 2 1 2 3 2 2 2 Output 0 2 6 Input 3 3 4 1 1 1 1 1 1 1 1 1 2 2 2 1 2 2 2 1 1 2 2 1 Output 6 3 3 Note Consider the first sample. The first query specifies pixel (2, 2), which is dark itself, so there are no valid light blocks, thus the answer is 0. The second query specifies pixel (1, 2). The biggest light block is the block with (1, 2) as its upper-left vertex and (1, 3) as its lower-right vertex. The last query specifies pixel (2, 2), which became light in the third operation. The biggest light block is the block with (1, 2) as its upper-left vertex and (3, 3) as its lower-right vertex. Submitted Solution: ``` n, m, q = [int(c) for c in input().split()] field = [] for i in range(n): field.append([int(c) for c in input().split()]) def max_dim(x, y, dx, dy): ix = x iy = y count = 0 while (0 <= ix < m) and (0 <= iy < n) and (field[iy][ix] == 1): ix += dx iy += dy count += 1 return count def find_square(arr, value): f = 0 while arr[f] < value: f += 1 l = len(arr) - 1 while arr[l] < value: l -= 1 return (l - f + 1) * value def count_one(x, y, dx, dy): x_ = x y_ = y max_y_ = n counter = [] for i in range(max_y_): counter.append(max_dim(x_, i, dx, dy)) for i in range(y_ - 1, -1, -1): if counter[i + 1] < counter[i]: counter[i] = counter[i + 1] for i in range(y_ + 1, max_y_): if counter[i - 1] < counter[i]: counter[i] = counter[i - 1] max = 0 for i in range(max_y_): t = find_square(counter, counter[i]) if t > max: max = t return max for i in range(q): order, x, y = [int(c) for c in input().split()] x -= 1 y -= 1 if order == 1: field[x][y] = (field[x][y] + 1) % 2 if order == 2: print(max(count_one(y, x, 1, 0),count_one(y, x, -1, 0))) ```
instruction
0
12,826
7
25,652
No
output
1
12,826
7
25,653
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Nanami is an expert at playing games. This day, Nanami's good friend Hajime invited her to watch a game of baseball. Unwilling as she was, she followed him to the stadium. But Nanami had no interest in the game, so she looked around to see if there was something that might interest her. That's when she saw the digital board at one end of the stadium. The digital board is n pixels in height and m pixels in width, every pixel is either light or dark. The pixels are described by its coordinate. The j-th pixel of the i-th line is pixel (i, j). The board displays messages by switching a combination of pixels to light, and the rest to dark. Nanami notices that the state of the pixels on the board changes from time to time. At certain times, certain pixels on the board may switch from light to dark, or from dark to light. Nanami wonders, what is the area of the biggest light block such that a specific pixel is on its side. A light block is a sub-rectangle of the board, in which all pixels are light. Pixel (i, j) belongs to a side of sub-rectangle with (x1, y1) and (x2, y2) as its upper-left and lower-right vertex if and only if it satisfies the logical condition: ((i = x1 or i = x2) and (y1 ≀ j ≀ y2)) or ((j = y1 or j = y2) and (x1 ≀ i ≀ x2)). Nanami has all the history of changing pixels, also she has some questions of the described type, can you answer them? Input The first line contains three space-separated integers n, m and q (1 ≀ n, m, q ≀ 1000) β€” the height and width of the digital board, and the number of operations. Then follow n lines, each line containing m space-separated integers. The j-th integer of the i-th line is ai, j β€” the initial state of pixel (i, j). * If ai, j = 0, pixel (i, j) is initially dark. * If ai, j = 1, pixel (i, j) is initially light. Then follow q lines, each line containing three space-separated integers op, x, and y (1 ≀ op ≀ 2; 1 ≀ x ≀ n; 1 ≀ y ≀ m), describing an operation. * If op = 1, the pixel at (x, y) changes its state (from light to dark or from dark to light). * If op = 2, Nanami queries the biggest light block with pixel (x, y) on its side. Output For each query, print a single line containing one integer β€” the answer to Nanami's query. Examples Input 3 4 5 0 1 1 0 1 0 0 1 0 1 1 0 2 2 2 2 1 2 1 2 2 1 2 3 2 2 2 Output 0 2 6 Input 3 3 4 1 1 1 1 1 1 1 1 1 2 2 2 1 2 2 2 1 1 2 2 1 Output 6 3 3 Note Consider the first sample. The first query specifies pixel (2, 2), which is dark itself, so there are no valid light blocks, thus the answer is 0. The second query specifies pixel (1, 2). The biggest light block is the block with (1, 2) as its upper-left vertex and (1, 3) as its lower-right vertex. The last query specifies pixel (2, 2), which became light in the third operation. The biggest light block is the block with (1, 2) as its upper-left vertex and (3, 3) as its lower-right vertex. Submitted Solution: ``` import sys read = lambda t=int: list(map(t,sys.stdin.readline().split())) array = lambda *ds: [array(*ds[1:]) for _ in range(ds[0])] if ds else 0 rots = [ (lambda x,y:(x,y)), (lambda x,y:(y,n-1-x)), (lambda x,y:(n-1-x,m-1-y)), (lambda x,y:(m-1-y,x))] invs = [ (lambda x,y:(x,y)), (lambda x,y:(n-1-y,x)), (lambda x,y:(n-1-x,m-1-y)), (lambda x,y:(y,m-1-x))] n, m, q = read() # print(rots[0](n-1,m-1)) # print(rots[1](n-1,m-1)) # print(rots[2](n-1,m-1)) # print(rots[3](n-1,m-1)) ar = [read() for _ in range(n)] cols = [array(n,m), array(m,n), array(n,m), array(m,n)] for i in range(4): for x in range(len(cols[i])): for y in range(len(cols[i][0])): ax, ay = invs[i](x,y) # print(i,x,y,ax,ay) cols[i][x][y] = ar[ax][ay] if x != 0 and cols[i][x][y] != 0: cols[i][x][y] += cols[i][x-1][y] # for ar in cols: # for row in ar: # print(row) # print() for _ in range(q): typ, x, y = read() x, y = x-1, y-1 if typ == 1: for i in range(4): rx, ry = rots[i](x, y) if cols[i][rx][ry]: cols[i][rx][ry] = 0 else: cols[i][rx][ry] = 1 if rx != 0: cols[i][rx][ry] += cols[i][rx-1][ry] while rx+1 != len(cols[i]) and cols[i][rx+1][ry]: cols[i][rx+1][ry] = cols[i][rx][ry]+1 rx += 1 if typ == 2: best = 0 for i in range(4): # for row in cols[i]: # print(row) rx, ry = rots[i](x, y) h = cols[i][rx][ry] a, b = ry-1, ry+1 while a != -1 or b != len(cols[i][0]): # print(a,b,i,len(cols[i][0])) if a != -1 and (b == len(cols[i][0]) or cols[i][rx][a] >= cols[i][rx][b]): h = min(h, cols[i][rx][a]) best = max(best, h*(b-a)) a -= 1 elif b != len(cols[i][0]) and (a == -1 or cols[i][rx][b] >= cols[i][rx][a]): h = min(h, cols[i][rx][b]) best = max(best, h*(b-a)) b += 1 print(best) # for ar in cols: # for row in ar: # print(row) # print() ```
instruction
0
12,827
7
25,654
No
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12,827
7
25,655
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Nanami is an expert at playing games. This day, Nanami's good friend Hajime invited her to watch a game of baseball. Unwilling as she was, she followed him to the stadium. But Nanami had no interest in the game, so she looked around to see if there was something that might interest her. That's when she saw the digital board at one end of the stadium. The digital board is n pixels in height and m pixels in width, every pixel is either light or dark. The pixels are described by its coordinate. The j-th pixel of the i-th line is pixel (i, j). The board displays messages by switching a combination of pixels to light, and the rest to dark. Nanami notices that the state of the pixels on the board changes from time to time. At certain times, certain pixels on the board may switch from light to dark, or from dark to light. Nanami wonders, what is the area of the biggest light block such that a specific pixel is on its side. A light block is a sub-rectangle of the board, in which all pixels are light. Pixel (i, j) belongs to a side of sub-rectangle with (x1, y1) and (x2, y2) as its upper-left and lower-right vertex if and only if it satisfies the logical condition: ((i = x1 or i = x2) and (y1 ≀ j ≀ y2)) or ((j = y1 or j = y2) and (x1 ≀ i ≀ x2)). Nanami has all the history of changing pixels, also she has some questions of the described type, can you answer them? Input The first line contains three space-separated integers n, m and q (1 ≀ n, m, q ≀ 1000) β€” the height and width of the digital board, and the number of operations. Then follow n lines, each line containing m space-separated integers. The j-th integer of the i-th line is ai, j β€” the initial state of pixel (i, j). * If ai, j = 0, pixel (i, j) is initially dark. * If ai, j = 1, pixel (i, j) is initially light. Then follow q lines, each line containing three space-separated integers op, x, and y (1 ≀ op ≀ 2; 1 ≀ x ≀ n; 1 ≀ y ≀ m), describing an operation. * If op = 1, the pixel at (x, y) changes its state (from light to dark or from dark to light). * If op = 2, Nanami queries the biggest light block with pixel (x, y) on its side. Output For each query, print a single line containing one integer β€” the answer to Nanami's query. Examples Input 3 4 5 0 1 1 0 1 0 0 1 0 1 1 0 2 2 2 2 1 2 1 2 2 1 2 3 2 2 2 Output 0 2 6 Input 3 3 4 1 1 1 1 1 1 1 1 1 2 2 2 1 2 2 2 1 1 2 2 1 Output 6 3 3 Note Consider the first sample. The first query specifies pixel (2, 2), which is dark itself, so there are no valid light blocks, thus the answer is 0. The second query specifies pixel (1, 2). The biggest light block is the block with (1, 2) as its upper-left vertex and (1, 3) as its lower-right vertex. The last query specifies pixel (2, 2), which became light in the third operation. The biggest light block is the block with (1, 2) as its upper-left vertex and (3, 3) as its lower-right vertex. Submitted Solution: ``` import sys f = sys.stdin n,m,q = map(int, f.readline().strip().split()) a = [] for i in range(n): a.append( [int(u) for u in f.readline().strip().split()] ) for i in range(q): op, y, x = map(int, f.readline().strip().split()) x -= 1 y -= 1 if op==1: a[y][x] = 1 - a[y][x] else: i = 0 max_x1 = m-1 while x+i<m: if a[y][x+i]==0: max_x1 = i-1 break max_x1 = i i += 1 i = 0 max_y1 = n-1 while y+i<n: if a[y+i][x]==0: max_y1 = i-1 break max_y1 = i i += 1 i = 0 min_x1 = 0 while x-i>=0: if a[y][x-i]==0: min_x1 = i-1 break min_x1 = i i += 1 i = 0 min_y1 = 0 while y-i>=0: if a[y-i][x]==0: min_y1 = i-1 break min_y1 = i i += 1 max_S = 0 #print(op, y, x, min_x1, max_x1, min_y1, max_y1) max_S = max(max_S, max_y1+min_y1+1) max_S = max(max_S, max_x1+min_x1+1) if max_x1>0: i = 1 up = min_y1 down = max_y1 while i<=max_x1 and (up>0 or down>0): for k in range(-up,down+1): if a[y+k][x+i]==0: if k<0: up = -k-1 else: down = k-1 max_S = max(max_S, (i+1)*(up+down+1)) i += 1 if max_y1>0: i = 1 up = min_x1 down = max_x1 while i<=max_y1 and (up>0 or down>0): for k in range(-up,down+1): if a[y+i][x+k]==0: if k<0: up = -k-1 else: down = k-1 max_S = max(max_S, (i+1)*(up+down+1)) i += 1 if min_x1>0: i = 1 up = min_y1 down = max_y1 while i<=min_x1 and (up>0 or down>0): for k in range(-up,down+1): if a[y+k][x-i]==0: if k<0: up = -k-1 else: down = k-1 max_S = max(max_S, (i+1)*(up+down+1)) i += 1 if min_y1>0: i = 1 up = min_x1 down = max_x1 while i<=min_y1 and (up>0 or down>0): for k in range(-up,down+1): if a[y-i][x+k]==0: if k<0: up = -k-1 else: down = k-1 max_S = max(max_S, (i+1)*(up+down+1)) i += 1 print(max_S) ```
instruction
0
12,828
7
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No
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12,828
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25,657
Provide a correct Python 3 solution for this coding contest problem. Ringo Kingdom Congress is voting on a bill. N members are present, and the i-th member (1 ≀ i ≀ N) has w_i white ballots and b_i blue ballots. Each member i will put all the w_i white ballots into the box if he/she is in favor of the bill, and put all the b_i blue ballots into the box if he/she is not in favor of the bill. No other action is allowed. For example, a member must not forfeit voting, or put only a part of his/her white ballots or a part of his/her blue ballots into the box. After all the members vote, if at least P percent of the ballots in the box is white, the bill is passed; if less than P percent of the ballots is white, the bill is rejected. In order for the bill to pass, at least how many members must be in favor of it? Constraints * 1 ≀ N ≀ 10^5 * 1 ≀ P ≀ 100 * 1 ≀ w_i ≀ 10^9 * 1 ≀ b_i ≀ 10^9 * All input values are integers. Input Input is given from Standard Input in the following format: N P w_1 b_1 w_2 b_2 : w_N b_N Output Print the minimum number of members in favor of the bill required for passage. Examples Input 4 75 1 1 1 1 1 1 1 1 Output 3 Input 4 75 1 1 1 1 1 1 100 1 Output 1 Input 5 60 6 3 5 9 3 4 7 8 4 7 Output 3
instruction
0
13,177
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"Correct Solution: ``` n, p = map(int, input().split()) e = [tuple(map(int, input().split())) for _ in range(n)] # e = [(w,b), ...] e2 = sorted(((100 - p) * w + p * b for w, b in e), reverse=True) rest = p * sum(b for _, b in e) cur = 0 while rest > 0 and cur < n: rest -= e2[cur] cur += 1 print(cur) ```
output
1
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Provide a correct Python 3 solution for this coding contest problem. Ringo Kingdom Congress is voting on a bill. N members are present, and the i-th member (1 ≀ i ≀ N) has w_i white ballots and b_i blue ballots. Each member i will put all the w_i white ballots into the box if he/she is in favor of the bill, and put all the b_i blue ballots into the box if he/she is not in favor of the bill. No other action is allowed. For example, a member must not forfeit voting, or put only a part of his/her white ballots or a part of his/her blue ballots into the box. After all the members vote, if at least P percent of the ballots in the box is white, the bill is passed; if less than P percent of the ballots is white, the bill is rejected. In order for the bill to pass, at least how many members must be in favor of it? Constraints * 1 ≀ N ≀ 10^5 * 1 ≀ P ≀ 100 * 1 ≀ w_i ≀ 10^9 * 1 ≀ b_i ≀ 10^9 * All input values are integers. Input Input is given from Standard Input in the following format: N P w_1 b_1 w_2 b_2 : w_N b_N Output Print the minimum number of members in favor of the bill required for passage. Examples Input 4 75 1 1 1 1 1 1 1 1 Output 3 Input 4 75 1 1 1 1 1 1 100 1 Output 1 Input 5 60 6 3 5 9 3 4 7 8 4 7 Output 3
instruction
0
13,178
7
26,356
"Correct Solution: ``` n, p = map(int, input().split()) w, b = [], [] for _ in range(n): ww, bb = map(int, input().split()) w.append(ww) b.append(bb) s = [] for ww, bb in zip(w, b): s.append((100 - p) * ww + p * bb) s.sort(reverse=True) score = -sum(b) * p cnt = 0 while score < 0: score += s[cnt] cnt += 1 print(cnt) # (100-p)*w-p*b>=0 ```
output
1
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Provide a correct Python 3 solution for this coding contest problem. Ringo Kingdom Congress is voting on a bill. N members are present, and the i-th member (1 ≀ i ≀ N) has w_i white ballots and b_i blue ballots. Each member i will put all the w_i white ballots into the box if he/she is in favor of the bill, and put all the b_i blue ballots into the box if he/she is not in favor of the bill. No other action is allowed. For example, a member must not forfeit voting, or put only a part of his/her white ballots or a part of his/her blue ballots into the box. After all the members vote, if at least P percent of the ballots in the box is white, the bill is passed; if less than P percent of the ballots is white, the bill is rejected. In order for the bill to pass, at least how many members must be in favor of it? Constraints * 1 ≀ N ≀ 10^5 * 1 ≀ P ≀ 100 * 1 ≀ w_i ≀ 10^9 * 1 ≀ b_i ≀ 10^9 * All input values are integers. Input Input is given from Standard Input in the following format: N P w_1 b_1 w_2 b_2 : w_N b_N Output Print the minimum number of members in favor of the bill required for passage. Examples Input 4 75 1 1 1 1 1 1 1 1 Output 3 Input 4 75 1 1 1 1 1 1 100 1 Output 1 Input 5 60 6 3 5 9 3 4 7 8 4 7 Output 3
instruction
0
13,179
7
26,358
"Correct Solution: ``` n,p=map(int,input().split()) l=[] for i in range(n): l.append(list(map(int,input().split()))) l.sort(key=lambda x:-(x[0]*(100-p)+x[1]*p)) z=0 for i in range(n): z+=-l[i][1]*p i=0 while z<0: z+=l[i][0]*(100-p)+l[i][1]*p i+=1 print(i) ```
output
1
13,179
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26,359
Provide a correct Python 3 solution for this coding contest problem. Ringo Kingdom Congress is voting on a bill. N members are present, and the i-th member (1 ≀ i ≀ N) has w_i white ballots and b_i blue ballots. Each member i will put all the w_i white ballots into the box if he/she is in favor of the bill, and put all the b_i blue ballots into the box if he/she is not in favor of the bill. No other action is allowed. For example, a member must not forfeit voting, or put only a part of his/her white ballots or a part of his/her blue ballots into the box. After all the members vote, if at least P percent of the ballots in the box is white, the bill is passed; if less than P percent of the ballots is white, the bill is rejected. In order for the bill to pass, at least how many members must be in favor of it? Constraints * 1 ≀ N ≀ 10^5 * 1 ≀ P ≀ 100 * 1 ≀ w_i ≀ 10^9 * 1 ≀ b_i ≀ 10^9 * All input values are integers. Input Input is given from Standard Input in the following format: N P w_1 b_1 w_2 b_2 : w_N b_N Output Print the minimum number of members in favor of the bill required for passage. Examples Input 4 75 1 1 1 1 1 1 1 1 Output 3 Input 4 75 1 1 1 1 1 1 100 1 Output 1 Input 5 60 6 3 5 9 3 4 7 8 4 7 Output 3
instruction
0
13,180
7
26,360
"Correct Solution: ``` N, P = map(int, input().split()) B = 0 Q = [] for i in range(N): w, b = map(int, input().split()) B += b Q.append((100 - P)*w + P*b) Q.sort() s = 0 for i in range(N): s += Q[-i-1] if B*P <= s: print(i+1) break ```
output
1
13,180
7
26,361
Provide a correct Python 3 solution for this coding contest problem. Ringo Kingdom Congress is voting on a bill. N members are present, and the i-th member (1 ≀ i ≀ N) has w_i white ballots and b_i blue ballots. Each member i will put all the w_i white ballots into the box if he/she is in favor of the bill, and put all the b_i blue ballots into the box if he/she is not in favor of the bill. No other action is allowed. For example, a member must not forfeit voting, or put only a part of his/her white ballots or a part of his/her blue ballots into the box. After all the members vote, if at least P percent of the ballots in the box is white, the bill is passed; if less than P percent of the ballots is white, the bill is rejected. In order for the bill to pass, at least how many members must be in favor of it? Constraints * 1 ≀ N ≀ 10^5 * 1 ≀ P ≀ 100 * 1 ≀ w_i ≀ 10^9 * 1 ≀ b_i ≀ 10^9 * All input values are integers. Input Input is given from Standard Input in the following format: N P w_1 b_1 w_2 b_2 : w_N b_N Output Print the minimum number of members in favor of the bill required for passage. Examples Input 4 75 1 1 1 1 1 1 1 1 Output 3 Input 4 75 1 1 1 1 1 1 100 1 Output 1 Input 5 60 6 3 5 9 3 4 7 8 4 7 Output 3
instruction
0
13,181
7
26,362
"Correct Solution: ``` N,P=[int(i) for i in input().split()] wb=[[int(j) for j in input().split()] for i in range(N)] xs = [(100-P)*w+P*b for w,b in wb] xs.sort() xs.reverse() score=0 for i in range(N): score-=P*wb[i][1] for i in range(N): score+=xs[i] if score>=0: print(i+1) break ```
output
1
13,181
7
26,363
Provide a correct Python 3 solution for this coding contest problem. Ringo Kingdom Congress is voting on a bill. N members are present, and the i-th member (1 ≀ i ≀ N) has w_i white ballots and b_i blue ballots. Each member i will put all the w_i white ballots into the box if he/she is in favor of the bill, and put all the b_i blue ballots into the box if he/she is not in favor of the bill. No other action is allowed. For example, a member must not forfeit voting, or put only a part of his/her white ballots or a part of his/her blue ballots into the box. After all the members vote, if at least P percent of the ballots in the box is white, the bill is passed; if less than P percent of the ballots is white, the bill is rejected. In order for the bill to pass, at least how many members must be in favor of it? Constraints * 1 ≀ N ≀ 10^5 * 1 ≀ P ≀ 100 * 1 ≀ w_i ≀ 10^9 * 1 ≀ b_i ≀ 10^9 * All input values are integers. Input Input is given from Standard Input in the following format: N P w_1 b_1 w_2 b_2 : w_N b_N Output Print the minimum number of members in favor of the bill required for passage. Examples Input 4 75 1 1 1 1 1 1 1 1 Output 3 Input 4 75 1 1 1 1 1 1 100 1 Output 1 Input 5 60 6 3 5 9 3 4 7 8 4 7 Output 3
instruction
0
13,182
7
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"Correct Solution: ``` from itertools import accumulate def solve(l, r): if l > r: return l m = (l + r) // 2 w, b = whites[m], blues[m] if w * 100 // (w + b) < p: l = m + 1 else: r = m - 1 return solve(l, r) n, p = map(int, input().split()) wbs = [tuple(map(int, input().split())) for _ in range(n)] ass = [(w * (100 - p) + b * p, i) for i, (w, b) in enumerate(wbs)] ass.sort(reverse=True) whites = list(accumulate(wbs[i][0] for _, i in ass)) blues = [wbs[i][1] for _, i in ass[1:]] blues = list(reversed(list(accumulate(reversed(blues))))) + [0] # print(list(whites)) # print(list(blues)) print(solve(0, n - 1) + 1) ```
output
1
13,182
7
26,365
Provide a correct Python 3 solution for this coding contest problem. Ringo Kingdom Congress is voting on a bill. N members are present, and the i-th member (1 ≀ i ≀ N) has w_i white ballots and b_i blue ballots. Each member i will put all the w_i white ballots into the box if he/she is in favor of the bill, and put all the b_i blue ballots into the box if he/she is not in favor of the bill. No other action is allowed. For example, a member must not forfeit voting, or put only a part of his/her white ballots or a part of his/her blue ballots into the box. After all the members vote, if at least P percent of the ballots in the box is white, the bill is passed; if less than P percent of the ballots is white, the bill is rejected. In order for the bill to pass, at least how many members must be in favor of it? Constraints * 1 ≀ N ≀ 10^5 * 1 ≀ P ≀ 100 * 1 ≀ w_i ≀ 10^9 * 1 ≀ b_i ≀ 10^9 * All input values are integers. Input Input is given from Standard Input in the following format: N P w_1 b_1 w_2 b_2 : w_N b_N Output Print the minimum number of members in favor of the bill required for passage. Examples Input 4 75 1 1 1 1 1 1 1 1 Output 3 Input 4 75 1 1 1 1 1 1 100 1 Output 1 Input 5 60 6 3 5 9 3 4 7 8 4 7 Output 3
instruction
0
13,183
7
26,366
"Correct Solution: ``` N, P = map(int, input().split()) xs = [] value = 0 for _ in range(N): w, b = map(int, input().split()) xs.append((100-P)*w+P*b) value -= P*b xs.sort(reverse=True) for i, x in enumerate(xs, 1): value += x if value >= 0: print(i) break ```
output
1
13,183
7
26,367
Provide tags and a correct Python 3 solution for this coding contest problem. Twilight Sparkle once got a crystal from the Crystal Mine. A crystal of size n (n is odd; n > 1) is an n Γ— n matrix with a diamond inscribed into it. You are given an odd integer n. You need to draw a crystal of size n. The diamond cells of the matrix should be represented by character "D". All other cells of the matrix should be represented by character "*". Look at the examples to understand what you need to draw. Input The only line contains an integer n (3 ≀ n ≀ 101; n is odd). Output Output a crystal of size n. Examples Input 3 Output *D* DDD *D* Input 5 Output **D** *DDD* DDDDD *DDD* **D** Input 7 Output ***D*** **DDD** *DDDDD* DDDDDDD *DDDDD* **DDD** ***D***
instruction
0
13,655
7
27,310
Tags: implementation Correct Solution: ``` n=int(input()) d=0 for i in range(n): if i<n//2: print("*"*(n//2-i)+"D"*(2*i+1)+"*"*(n//2-i)) elif i==n//2: print("D"*n) elif i>n//2: print("*"*(i-n//2)+"D"*(n-2-2*d)+"*"*(n//2-(n-i-1))) d+=1 ```
output
1
13,655
7
27,311
Provide tags and a correct Python 3 solution for this coding contest problem. Twilight Sparkle once got a crystal from the Crystal Mine. A crystal of size n (n is odd; n > 1) is an n Γ— n matrix with a diamond inscribed into it. You are given an odd integer n. You need to draw a crystal of size n. The diamond cells of the matrix should be represented by character "D". All other cells of the matrix should be represented by character "*". Look at the examples to understand what you need to draw. Input The only line contains an integer n (3 ≀ n ≀ 101; n is odd). Output Output a crystal of size n. Examples Input 3 Output *D* DDD *D* Input 5 Output **D** *DDD* DDDDD *DDD* **D** Input 7 Output ***D*** **DDD** *DDDDD* DDDDDDD *DDDDD* **DDD** ***D***
instruction
0
13,656
7
27,312
Tags: implementation Correct Solution: ``` import math, os, sys import string, re import itertools, functools, operator from collections import Counter def inputint(): return int(input()) def inputarray(func=int): return map(func, input().split()) n = inputint() for i in range(n): x = abs(n//2 - i) print('*'*x + 'D'*(n - 2*x) + '*'*x) ```
output
1
13,656
7
27,313
Provide tags and a correct Python 3 solution for this coding contest problem. Twilight Sparkle once got a crystal from the Crystal Mine. A crystal of size n (n is odd; n > 1) is an n Γ— n matrix with a diamond inscribed into it. You are given an odd integer n. You need to draw a crystal of size n. The diamond cells of the matrix should be represented by character "D". All other cells of the matrix should be represented by character "*". Look at the examples to understand what you need to draw. Input The only line contains an integer n (3 ≀ n ≀ 101; n is odd). Output Output a crystal of size n. Examples Input 3 Output *D* DDD *D* Input 5 Output **D** *DDD* DDDDD *DDD* **D** Input 7 Output ***D*** **DDD** *DDDDD* DDDDDDD *DDDDD* **DDD** ***D***
instruction
0
13,657
7
27,314
Tags: implementation Correct Solution: ``` n = int(input()) k = n // 2 ans = [['*']*n for i in range(n)] i = 0 while k >= 0: for j in range(k,n-k): ans[i][j] = 'D' k -= 1 i += 1 k = 1 while k <=n // 2: for j in range(k,n-k): ans[i][j] = 'D' k += 1 i += 1 for i in range(n): for j in range(n): print(ans[i][j], end = '') print('') ```
output
1
13,657
7
27,315
Provide tags and a correct Python 3 solution for this coding contest problem. Twilight Sparkle once got a crystal from the Crystal Mine. A crystal of size n (n is odd; n > 1) is an n Γ— n matrix with a diamond inscribed into it. You are given an odd integer n. You need to draw a crystal of size n. The diamond cells of the matrix should be represented by character "D". All other cells of the matrix should be represented by character "*". Look at the examples to understand what you need to draw. Input The only line contains an integer n (3 ≀ n ≀ 101; n is odd). Output Output a crystal of size n. Examples Input 3 Output *D* DDD *D* Input 5 Output **D** *DDD* DDDDD *DDD* **D** Input 7 Output ***D*** **DDD** *DDDDD* DDDDDDD *DDDDD* **DDD** ***D***
instruction
0
13,658
7
27,316
Tags: implementation Correct Solution: ``` n=int(input()) x='' for a in range(1,int((n+3)/2)): b=(n+1-2*a)/2 x+=int(b)*'*' x+=(2*a-1)*'D' x+=int(b)*'*' print(x) x='' for a in range(1,int((n+1)/2)): b=n-2*a x+=a*'*' x+=b*'D' x+=a*'*' print(x) x='' ```
output
1
13,658
7
27,317
Provide tags and a correct Python 3 solution for this coding contest problem. Twilight Sparkle once got a crystal from the Crystal Mine. A crystal of size n (n is odd; n > 1) is an n Γ— n matrix with a diamond inscribed into it. You are given an odd integer n. You need to draw a crystal of size n. The diamond cells of the matrix should be represented by character "D". All other cells of the matrix should be represented by character "*". Look at the examples to understand what you need to draw. Input The only line contains an integer n (3 ≀ n ≀ 101; n is odd). Output Output a crystal of size n. Examples Input 3 Output *D* DDD *D* Input 5 Output **D** *DDD* DDDDD *DDD* **D** Input 7 Output ***D*** **DDD** *DDDDD* DDDDDDD *DDDDD* **DDD** ***D***
instruction
0
13,659
7
27,318
Tags: implementation Correct Solution: ``` n = int(input()); m = n//2; k = 1 for i in range(n): if k<n: print('*'*m + 'D'*k + '*'*m) m-=1; k+=2 k=n for i in range(n): print('*'*m + 'D'*k + '*'*m) m+=1; k-=2 if m>n//2: break ```
output
1
13,659
7
27,319
Provide tags and a correct Python 3 solution for this coding contest problem. Twilight Sparkle once got a crystal from the Crystal Mine. A crystal of size n (n is odd; n > 1) is an n Γ— n matrix with a diamond inscribed into it. You are given an odd integer n. You need to draw a crystal of size n. The diamond cells of the matrix should be represented by character "D". All other cells of the matrix should be represented by character "*". Look at the examples to understand what you need to draw. Input The only line contains an integer n (3 ≀ n ≀ 101; n is odd). Output Output a crystal of size n. Examples Input 3 Output *D* DDD *D* Input 5 Output **D** *DDD* DDDDD *DDD* **D** Input 7 Output ***D*** **DDD** *DDDDD* DDDDDDD *DDDDD* **DDD** ***D***
instruction
0
13,660
7
27,320
Tags: implementation Correct Solution: ``` a = int(input()) mat=[] sat=[] fat=[] rat=[] for i in range(a//2+1): for j in range(a): if i+j>=a//2 and i+j<=((a//2)+2*i): sat.append('D') else: sat.append('*') mat.append(sat) sat=[] for i in mat: for j in i: print(j,end='') print() for i in range(1,a//2+1): for j in range(a): print(mat[a//2-i][j],end='') print() ```
output
1
13,660
7
27,321
Provide tags and a correct Python 3 solution for this coding contest problem. Twilight Sparkle once got a crystal from the Crystal Mine. A crystal of size n (n is odd; n > 1) is an n Γ— n matrix with a diamond inscribed into it. You are given an odd integer n. You need to draw a crystal of size n. The diamond cells of the matrix should be represented by character "D". All other cells of the matrix should be represented by character "*". Look at the examples to understand what you need to draw. Input The only line contains an integer n (3 ≀ n ≀ 101; n is odd). Output Output a crystal of size n. Examples Input 3 Output *D* DDD *D* Input 5 Output **D** *DDD* DDDDD *DDD* **D** Input 7 Output ***D*** **DDD** *DDDDD* DDDDDDD *DDDDD* **DDD** ***D***
instruction
0
13,661
7
27,322
Tags: implementation Correct Solution: ``` n=int(input()) a=n//2 c=0 for i in range((n//2)+1): for j in range(0,a): print("*",end="") for k in range(0,c+1): print("D",end="") c=c+2 for j in range(0,a): print("*",end="") print() a=a-1 c=n-2 b=1 for i in range(n//2): for j in range(0,b): print("*",end="") for k in range(0,c): print("D",end="") c=c-2 for j in range(0,b): print("*",end="") print() b=b+1 ```
output
1
13,661
7
27,323
Provide tags and a correct Python 3 solution for this coding contest problem. Twilight Sparkle once got a crystal from the Crystal Mine. A crystal of size n (n is odd; n > 1) is an n Γ— n matrix with a diamond inscribed into it. You are given an odd integer n. You need to draw a crystal of size n. The diamond cells of the matrix should be represented by character "D". All other cells of the matrix should be represented by character "*". Look at the examples to understand what you need to draw. Input The only line contains an integer n (3 ≀ n ≀ 101; n is odd). Output Output a crystal of size n. Examples Input 3 Output *D* DDD *D* Input 5 Output **D** *DDD* DDDDD *DDD* **D** Input 7 Output ***D*** **DDD** *DDDDD* DDDDDDD *DDDDD* **DDD** ***D***
instruction
0
13,662
7
27,324
Tags: implementation Correct Solution: ``` n = int(input()) add = -1 a = n // 2 for i in range(n): if a == -1: a = 1 add *= -1 print("*" * a + "D" * (n - a * 2) + "*" * a) a += add ```
output
1
13,662
7
27,325
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Twilight Sparkle once got a crystal from the Crystal Mine. A crystal of size n (n is odd; n > 1) is an n Γ— n matrix with a diamond inscribed into it. You are given an odd integer n. You need to draw a crystal of size n. The diamond cells of the matrix should be represented by character "D". All other cells of the matrix should be represented by character "*". Look at the examples to understand what you need to draw. Input The only line contains an integer n (3 ≀ n ≀ 101; n is odd). Output Output a crystal of size n. Examples Input 3 Output *D* DDD *D* Input 5 Output **D** *DDD* DDDDD *DDD* **D** Input 7 Output ***D*** **DDD** *DDDDD* DDDDDDD *DDDDD* **DDD** ***D*** Submitted Solution: ``` n = int(input()) for i in range(n): buf = abs((n-1)//2-i) other = n-2*buf print('*'*buf+'D'*other+'*'*buf) ```
instruction
0
13,663
7
27,326
Yes
output
1
13,663
7
27,327
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Twilight Sparkle once got a crystal from the Crystal Mine. A crystal of size n (n is odd; n > 1) is an n Γ— n matrix with a diamond inscribed into it. You are given an odd integer n. You need to draw a crystal of size n. The diamond cells of the matrix should be represented by character "D". All other cells of the matrix should be represented by character "*". Look at the examples to understand what you need to draw. Input The only line contains an integer n (3 ≀ n ≀ 101; n is odd). Output Output a crystal of size n. Examples Input 3 Output *D* DDD *D* Input 5 Output **D** *DDD* DDDDD *DDD* **D** Input 7 Output ***D*** **DDD** *DDDDD* DDDDDDD *DDDDD* **DDD** ***D*** Submitted Solution: ``` height = int(input()) diamonds_on_line=1 while diamonds_on_line!=height: print(((height-diamonds_on_line)//2)*'*'+diamonds_on_line*'D'+((height-diamonds_on_line)//2)*'*') diamonds_on_line+=2 print(diamonds_on_line*'D') while diamonds_on_line!=1: diamonds_on_line-=2 print(((height-diamonds_on_line)//2)*'*'+diamonds_on_line*'D'+((height-diamonds_on_line)//2)*'*') ```
instruction
0
13,664
7
27,328
Yes
output
1
13,664
7
27,329
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Twilight Sparkle once got a crystal from the Crystal Mine. A crystal of size n (n is odd; n > 1) is an n Γ— n matrix with a diamond inscribed into it. You are given an odd integer n. You need to draw a crystal of size n. The diamond cells of the matrix should be represented by character "D". All other cells of the matrix should be represented by character "*". Look at the examples to understand what you need to draw. Input The only line contains an integer n (3 ≀ n ≀ 101; n is odd). Output Output a crystal of size n. Examples Input 3 Output *D* DDD *D* Input 5 Output **D** *DDD* DDDDD *DDD* **D** Input 7 Output ***D*** **DDD** *DDDDD* DDDDDDD *DDDDD* **DDD** ***D*** Submitted Solution: ``` n=int(input()) k=[] s='' a=n-1 a=a//2 b=1 p=(n//2)+1 for i in range(p): for j in range(a): s+='*' for j in range(b): s+='D' for j in range(a): s+='*' k.append(s) s='' b=b+2 a=a-1 s='' a=1 b=n-2 for i in range(p-1): for j in range(a): s+='*' for j in range(b): s+='D' for j in range(a): s+='*' k.append(s) s='' b=b-2 a=a+1 for i in k: print(i) ```
instruction
0
13,665
7
27,330
Yes
output
1
13,665
7
27,331
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Twilight Sparkle once got a crystal from the Crystal Mine. A crystal of size n (n is odd; n > 1) is an n Γ— n matrix with a diamond inscribed into it. You are given an odd integer n. You need to draw a crystal of size n. The diamond cells of the matrix should be represented by character "D". All other cells of the matrix should be represented by character "*". Look at the examples to understand what you need to draw. Input The only line contains an integer n (3 ≀ n ≀ 101; n is odd). Output Output a crystal of size n. Examples Input 3 Output *D* DDD *D* Input 5 Output **D** *DDD* DDDDD *DDD* **D** Input 7 Output ***D*** **DDD** *DDDDD* DDDDDDD *DDDDD* **DDD** ***D*** Submitted Solution: ``` #####--------------Template Begin-------------------#### import math import sys import string #input = sys.stdin.readline def ri(): #Regular input return input() def ii(): #integer input return int(input()) def li(): #list input return input().split() def mi(): #map input return list(map(int, input().split())) #####---------------Template Ends-------------------###### n=ii() for i in range(1, n-1, 2): print(("D"*i).center(n, "*")) print("D"*n) for i in range(n-2,0,-2): print(("D"*i).center(n, "*")) ```
instruction
0
13,666
7
27,332
Yes
output
1
13,666
7
27,333
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Twilight Sparkle once got a crystal from the Crystal Mine. A crystal of size n (n is odd; n > 1) is an n Γ— n matrix with a diamond inscribed into it. You are given an odd integer n. You need to draw a crystal of size n. The diamond cells of the matrix should be represented by character "D". All other cells of the matrix should be represented by character "*". Look at the examples to understand what you need to draw. Input The only line contains an integer n (3 ≀ n ≀ 101; n is odd). Output Output a crystal of size n. Examples Input 3 Output *D* DDD *D* Input 5 Output **D** *DDD* DDDDD *DDD* **D** Input 7 Output ***D*** **DDD** *DDDDD* DDDDDDD *DDDDD* **DDD** ***D*** Submitted Solution: ``` def solve(): n = int(input()) score = n // 2 lst = [] for i in range(score, -1, -1): string = "*" * i + "D" * (n - 2*i) + "*" * i lst.append(string) for j in lst: print(j) for k in lst[1::-1]: print(k) solve() ```
instruction
0
13,667
7
27,334
No
output
1
13,667
7
27,335
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Twilight Sparkle once got a crystal from the Crystal Mine. A crystal of size n (n is odd; n > 1) is an n Γ— n matrix with a diamond inscribed into it. You are given an odd integer n. You need to draw a crystal of size n. The diamond cells of the matrix should be represented by character "D". All other cells of the matrix should be represented by character "*". Look at the examples to understand what you need to draw. Input The only line contains an integer n (3 ≀ n ≀ 101; n is odd). Output Output a crystal of size n. Examples Input 3 Output *D* DDD *D* Input 5 Output **D** *DDD* DDDDD *DDD* **D** Input 7 Output ***D*** **DDD** *DDDDD* DDDDDDD *DDDDD* **DDD** ***D*** Submitted Solution: ``` n = int(input()) c = "*" *n x = 1 mid = False for i in range(n): if i == (n-1)/2: mid = True a = c[:int((n-x)/2)+1] b = "D"*x print(a+b+a) if mid == True: x -= 2 else: x += 2 ```
instruction
0
13,668
7
27,336
No
output
1
13,668
7
27,337
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Twilight Sparkle once got a crystal from the Crystal Mine. A crystal of size n (n is odd; n > 1) is an n Γ— n matrix with a diamond inscribed into it. You are given an odd integer n. You need to draw a crystal of size n. The diamond cells of the matrix should be represented by character "D". All other cells of the matrix should be represented by character "*". Look at the examples to understand what you need to draw. Input The only line contains an integer n (3 ≀ n ≀ 101; n is odd). Output Output a crystal of size n. Examples Input 3 Output *D* DDD *D* Input 5 Output **D** *DDD* DDDDD *DDD* **D** Input 7 Output ***D*** **DDD** *DDDDD* DDDDDDD *DDDDD* **DDD** ***D*** Submitted Solution: ``` from __future__ import print_function print ("self") y= input("please enter number:") x=int(y) count = 1 side = True for i in range(x): for j in range(int((x-count)/2)): print('*',end='') for j in range(count): print('D',end='') for j in range(int((x-count)/2)): print('*',end='') if(i!=x-1): print("") if(count>=x): side =False if(side): count+=2 else: count-=2 ```
instruction
0
13,669
7
27,338
No
output
1
13,669
7
27,339